🧠:<think>Alright, so I've got this problem here: Given a positive odd number ( m ) and an integer ( a ), I need to prove that for any real number ( c ), the number of integers ( x ) in the interval ([c, c + sqrt{m}]) such that ( x^2 equiv a pmod{m} ) is at most ( 2 + log_2 m ). Hmm, okay. Let me break this down. First, ( m ) is a positive odd integer, and ( a ) is just some integer. The interval ([c, c + sqrt{m}]) is a range of real numbers, but we're only interested in the integers within this interval. For each integer ( x ) in this interval, we're checking if ( x^2 ) is congruent to ( a ) modulo ( m ). The goal is to show that the number of such integers ( x ) doesn't exceed ( 2 + log_2 m ). I think the first step is to understand what ( x^2 equiv a pmod{m} ) means. This congruence implies that ( m ) divides ( x^2 - a ), right? So, ( x^2 - a ) is a multiple of ( m ). That is, ( x^2 = a + km ) for some integer ( k ). Now, since ( m ) is odd, it can be factored into prime powers. Maybe I can use properties of quadratic residues modulo primes or prime powers? I remember that for a prime ( p ), the number of solutions to ( x^2 equiv a pmod{p} ) is at most 2. If ( a ) is a quadratic residue modulo ( p ), there are two solutions; otherwise, there are none. But here, we're dealing with an interval of length ( sqrt{m} ). So, the interval ([c, c + sqrt{m}]) can contain up to ( sqrt{m} ) integers. But we're not counting all integers, only those ( x ) such that ( x^2 equiv a pmod{m} ). I wonder if the Chinese Remainder Theorem can help here since ( m ) is a product of primes. If ( m = p_1^{e_1} p_2^{e_2} dots p_k^{e_k} ), then solving ( x^2 equiv a pmod{m} ) is equivalent to solving the system of congruences ( x^2 equiv a pmod{p_i^{e_i}} ) for each ( i ). For each prime power ( p_i^{e_i} ), the number of solutions to ( x^2 equiv a pmod{p_i^{e_i}} ) is either 0 or 2, assuming ( a ) is a quadratic residue modulo ( p_i^{e_i} ). So, if ( a ) is a quadratic residue modulo each prime power dividing ( m ), then the total number of solutions modulo ( m ) is ( 2^k ), where ( k ) is the number of distinct prime factors of ( m ). Wait, but ( m ) is odd, so all its prime factors are odd. Each quadratic congruence modulo ( p_i^{e_i} ) can have at most 2 solutions. So, if ( a ) is a quadratic residue modulo each ( p_i^{e_i} ), the total number of solutions modulo ( m ) is ( 2^k ). But how does this relate to the number of solutions within an interval of length ( sqrt{m} )? Maybe I can use some form of the pigeonhole principle or spacing of solutions. Let me consider the spacing between solutions. If ( x ) is a solution, then another solution ( x' ) must satisfy ( x' equiv x pmod{m} ) or ( x' equiv -x pmod{m} ). But within an interval of length ( sqrt{m} ), how many such ( x ) can there be? Alternatively, maybe I can use the fact that the number of solutions to ( x^2 equiv a pmod{m} ) is bounded by the number of prime factors of ( m ). Since each prime power contributes at most 2 solutions, the total number is ( 2^k ). But we need a bound in terms of ( log_2 m ). Since ( m ) is a product of primes, the number of prime factors ( k ) satisfies ( 2^k leq m ), so ( k leq log_2 m ). Wait, that might be the key. If ( m ) has at most ( log_2 m ) distinct prime factors, then the number of solutions to ( x^2 equiv a pmod{m} ) is at most ( 2^{log_2 m} = m ). But that doesn't help because we have an interval of length ( sqrt{m} ), and ( m ) is larger than ( sqrt{m} ). Hmm, perhaps I'm approaching this the wrong way. Maybe I should consider the interval length ( sqrt{m} ) and how many solutions can lie within that interval. Since the solutions are periodic modulo ( m ), the maximum number of solutions in any interval of length ( sqrt{m} ) should be related to the number of solutions modulo ( m ). But I'm not sure. Maybe I should look at a specific example. Let me take ( m = 9 ), which is ( 3^2 ). Then ( sqrt{9} = 3 ). Let's say ( a = 0 ). Then ( x^2 equiv 0 pmod{9} ) implies ( x equiv 0 pmod{3} ). In any interval of length 3, how many multiples of 3 are there? At most 2, right? For example, in [1,4], the multiples are 3. In [2,5], multiples are 3. In [0,3], multiples are 0, 3. So, maximum 2. But if ( a = 1 ), then ( x^2 equiv 1 pmod{9} ) has solutions ( x equiv 1 ) and ( x equiv 8 pmod{9} ). So in any interval of length 3, how many solutions can there be? Let's take [1,4]: 1 and 8 are not in this interval. [2,5]: 8 is not in this interval. [0,3]: only 1 is in the interval. [4,7]: 8 is not in this interval. [5,8]: 8 is in the interval. So, maximum 1 solution in any interval of length 3. Wait, but according to the problem statement, the bound is ( 2 + log_2 m ). For ( m = 9 ), ( log_2 9 approx 3.17 ), so ( 2 + 3.17 approx 5.17 ). But in reality, the maximum number of solutions is 2. So, the bound is not tight, but it's an upper bound. Another example: ( m = 15 ), which factors into 3 and 5. So, ( sqrt{15} approx 3.87 ). Let's take ( a = 1 ). Then ( x^2 equiv 1 pmod{15} ) has solutions ( x equiv 1, 4, 11, 14 pmod{15} ). So, 4 solutions modulo 15. In an interval of length approximately 3.87, how many solutions can there be? Let's take [0,3.87]. The solutions in this interval are 1 and 4 is outside. Wait, 4 is greater than 3.87, so only 1. Similarly, in [1,4.87], solutions are 1 and 4. So, 2 solutions. Thus, even though there are 4 solutions modulo 15, in any interval of length ( sqrt{15} ), we have at most 2 solutions. So, the bound ( 2 + log_2 15 approx 2 + 3.907 = 5.907 ) is still an upper bound. Hmm, so maybe the number of solutions in any interval of length ( sqrt{m} ) is related to the number of prime factors of ( m ). Since each prime factor can contribute at most 2 solutions, and the interval can contain at most one solution per prime factor? Not sure. Wait, maybe it's about the number of possible residues. Since ( m ) is odd, and we're looking at ( x ) in an interval of length ( sqrt{m} ), the number of possible residues modulo ( m ) is ( m ), but the interval only contains ( sqrt{m} ) integers. So, the number of possible residues is much larger than the number of integers in the interval. But how does that relate to the number of solutions? Each solution corresponds to a residue class modulo ( m ) where ( x^2 equiv a pmod{m} ). So, if there are ( t ) such residue classes, how many can lie within an interval of length ( sqrt{m} )?I think I need to use some form of the Chinese Remainder Theorem or maybe look into the distribution of quadratic residues modulo ( m ). Alternatively, maybe I can use the fact that the number of solutions to ( x^2 equiv a pmod{m} ) is bounded by the number of prime factors of ( m ). Since each prime power contributes at most 2 solutions, the total number of solutions modulo ( m ) is ( 2^k ), where ( k ) is the number of distinct prime factors of ( m ). But wait, ( k leq log_2 m ) because ( m geq 2^k ). So, the number of solutions is at most ( 2^{log_2 m} = m ), which is not helpful. Wait, no. Actually, if ( m ) has ( k ) distinct prime factors, then ( m geq 2^k ), so ( k leq log_2 m ). Therefore, the number of solutions to ( x^2 equiv a pmod{m} ) is at most ( 2^k leq 2^{log_2 m} = m ). But again, this doesn't help because the interval is of length ( sqrt{m} ). Maybe I need to consider the spacing of solutions. If ( x ) is a solution, then the next solution would be at least ( m ) apart, right? Because solutions are periodic modulo ( m ). So, within an interval of length ( sqrt{m} ), there can be at most one solution from each residue class. But wait, that's not necessarily true because if ( m ) is composite, there could be multiple solutions within the interval. Alternatively, perhaps I can use the fact that the number of solutions in the interval is related to the number of prime factors. Since each prime factor can contribute at most 2 solutions, and the interval can contain at most one solution from each prime factor, the total number is bounded by ( 2k ), where ( k ) is the number of prime factors. But ( k leq log_2 m ), so the total number is ( 2 log_2 m ). But the problem states ( 2 + log_2 m ), which is a bit different. Wait, maybe I'm overcomplicating it. Let me think differently. Suppose I have an interval of length ( sqrt{m} ). The number of integers in this interval is at most ( sqrt{m} + 1 ). For each integer ( x ) in this interval, we're checking if ( x^2 equiv a pmod{m} ). Now, consider the function ( f(x) = x^2 pmod{m} ). The number of distinct values of ( f(x) ) as ( x ) ranges over the integers is ( m ). But in our interval, we have ( sqrt{m} ) integers, so the number of distinct residues ( f(x) ) can take is at most ( sqrt{m} ). However, we're interested in how many times ( f(x) = a ). If ( a ) is a quadratic residue modulo ( m ), then there are ( 2^k ) solutions, but in our interval, we can have at most one solution from each residue class. Wait, maybe I need to use the fact that the number of solutions to ( x^2 equiv a pmod{m} ) is bounded by the number of prime factors. Since each prime power contributes at most 2 solutions, the total number of solutions modulo ( m ) is ( 2^k ), where ( k ) is the number of distinct prime factors. But how does this translate to the number of solutions in the interval? Since the interval is of length ( sqrt{m} ), and the solutions are spread out every ( m ), the number of solutions in the interval should be related to the number of solutions modulo ( m ). Wait, maybe I can use the fact that the number of solutions in any interval of length ( L ) is at most ( L / m times ) the total number of solutions modulo ( m ). But that seems vague. Alternatively, perhaps I can use the fact that the number of solutions in the interval is at most the number of solutions modulo ( m ) divided by the interval length. But that doesn't make sense because the interval length is ( sqrt{m} ), which is smaller than ( m ). I'm getting stuck here. Maybe I need to look up some related theorems or properties. I recall that in analytic number theory, there are bounds on the number of solutions to congruences in short intervals. Maybe something like that can be applied here. Wait, I think I remember something called the "Bombieri-Vinogradov theorem," but that's more about primes in arithmetic progressions. Maybe not directly applicable. Alternatively, perhaps I can use the fact that the number of solutions to ( x^2 equiv a pmod{m} ) is bounded by the number of prime factors of ( m ). Since each prime power contributes at most 2 solutions, the total number of solutions modulo ( m ) is ( 2^k ), where ( k ) is the number of distinct prime factors. But again, how does this relate to the interval? Maybe the key is that within any interval of length ( sqrt{m} ), the number of solutions is bounded by the number of prime factors. Since each prime factor can contribute at most 2 solutions, and the interval can contain at most one solution from each prime factor, the total number is ( 2k ). But ( k leq log_2 m ), so the total number is ( 2 log_2 m ). But the problem statement has ( 2 + log_2 m ), which is slightly different. Maybe there's a constant factor involved. Wait, maybe the factor of 2 comes from the fact that each prime can contribute 2 solutions, but in the interval, we can have at most one solution from each pair of solutions. So, it's ( log_2 m ) plus 2. I'm not entirely sure, but perhaps I can try to formalize this. Suppose ( m ) has ( k ) distinct prime factors. Then, the number of solutions to ( x^2 equiv a pmod{m} ) is at most ( 2^k ). However, in any interval of length ( sqrt{m} ), the number of solutions is at most ( k + 1 ), because each prime can contribute at most one solution in the interval. But that doesn't align with the bound given. Wait, maybe I need to consider that each prime can contribute at most 2 solutions, but within the interval, you can have at most one solution per prime. So, the total number is ( 2k ). But ( k leq log_2 m ), so ( 2k leq 2 log_2 m ). But the problem states ( 2 + log_2 m ), which is a bit different. Maybe there's a different approach. Perhaps I can use the fact that the number of solutions in the interval is bounded by the number of prime factors plus 2. Since each prime can contribute at most one solution, and there are two trivial solutions (if any), the total is ( 2 + log_2 m ). I think I need to consider the structure of the solutions. If ( m ) is a prime power, then there are at most 2 solutions. If ( m ) is composite, the number of solutions is multiplicative. So, if ( m = p_1^{e_1} p_2^{e_2} dots p_k^{e_k} ), then the number of solutions modulo ( m ) is ( 2^k ). But in the interval, how many of these can lie within ( sqrt{m} )? Since the solutions are spread out every ( m ), the number in the interval is roughly ( sqrt{m} / m = 1 / sqrt{m} times ) total solutions. But that seems too small. Wait, no. The interval is of length ( sqrt{m} ), so the number of solutions is roughly the total number of solutions multiplied by ( sqrt{m} / m = 1 / sqrt{m} ). But since the total number of solutions is ( 2^k ), the number in the interval is ( 2^k / sqrt{m} ). But ( 2^k leq m ), so ( 2^k / sqrt{m} leq sqrt{m} ). But that's not helpful because ( sqrt{m} ) is larger than ( 2 + log_2 m ) for large ( m ). Hmm, I'm going in circles here. Maybe I need to look at the problem differently. Let's think about the fact that ( x ) and ( x + sqrt{m} ) are in the interval. If ( x ) is a solution, then ( x + sqrt{m} ) would be another solution if ( (x + sqrt{m})^2 equiv a pmod{m} ). But expanding that, ( x^2 + 2xsqrt{m} + m equiv a pmod{m} ). Since ( m equiv 0 pmod{m} ), this simplifies to ( x^2 + 2xsqrt{m} equiv a pmod{m} ). But ( x^2 equiv a pmod{m} ), so ( 2xsqrt{m} equiv 0 pmod{m} ). This implies ( 2xsqrt{m} ) is a multiple of ( m ), so ( 2xsqrt{m} = km ) for some integer ( k ). Simplifying, ( x = k sqrt{m} / 2 ). But ( x ) is an integer, so ( sqrt{m} ) must divide ( 2x ). Since ( m ) is odd, ( sqrt{m} ) is either integer or irrational. Wait, ( m ) is a positive odd integer, so ( sqrt{m} ) is either integer or irrational. If ( sqrt{m} ) is integer, then ( m ) is a perfect square. In that case, ( x = k sqrt{m} / 2 ). For ( x ) to be integer, ( k ) must be even if ( sqrt{m} ) is odd. But since ( m ) is odd, ( sqrt{m} ) is also odd. So, ( k ) must be even, say ( k = 2l ). Then, ( x = l sqrt{m} ). But ( x ) is in the interval ([c, c + sqrt{m}]), so ( l sqrt{m} ) must be in this interval. But ( l sqrt{m} ) is much larger than ( sqrt{m} ) unless ( l = 0 ). So, ( x = 0 ) is a solution only if ( 0 ) is in the interval. But the interval is arbitrary, so this doesn't help. If ( sqrt{m} ) is irrational, then ( x ) cannot be an integer unless ( k = 0 ), which gives ( x = 0 ). Again, not helpful. So, this approach doesn't seem to work. Maybe I need to consider the multiplicative structure of ( m ). Since ( m ) is odd, it can be factored into primes, and by the Chinese Remainder Theorem, the solutions modulo ( m ) correspond to solutions modulo each prime power dividing ( m ). For each prime power ( p^e ), there are at most 2 solutions to ( x^2 equiv a pmod{p^e} ). Therefore, the total number of solutions modulo ( m ) is ( 2^k ), where ( k ) is the number of distinct prime factors of ( m ). Now, within the interval ([c, c + sqrt{m}]), the number of solutions is at most the number of solutions modulo ( m ) divided by the length of the interval. But that seems vague. Wait, maybe I can use the fact that the solutions are spaced at least ( m ) apart. So, in an interval of length ( sqrt{m} ), there can be at most one solution from each residue class. But if there are ( 2^k ) residue classes, then the number of solutions in the interval is at most ( 2^k times (sqrt{m} / m) = 2^k / sqrt{m} ). But ( 2^k leq m ), so ( 2^k / sqrt{m} leq sqrt{m} ). Again, not helpful. I think I need to approach this differently. Maybe consider the fact that the number of solutions in the interval is related to the number of prime factors and the interval length. Let me think about the multiplicative functions. The number of solutions to ( x^2 equiv a pmod{m} ) is multiplicative, so if ( m = m_1 m_2 ) with ( gcd(m_1, m_2) = 1 ), then the number of solutions modulo ( m ) is the product of the number of solutions modulo ( m_1 ) and ( m_2 ). Since each prime power contributes at most 2 solutions, the total number is ( 2^k ). Now, in the interval, how many of these can lie? Maybe the number is bounded by ( 2k ), because for each prime, you can have at most 2 solutions, but in the interval, you can have at most one from each pair. But ( k leq log_2 m ), so ( 2k leq 2 log_2 m ). But the problem states ( 2 + log_2 m ). Maybe there's a better way to bound it. Wait, perhaps the number of solutions in the interval is at most the number of prime factors plus 2. Because you have the trivial solutions and then one per prime factor. So, ( 2 + k ), and since ( k leq log_2 m ), the total is ( 2 + log_2 m ). I think that makes sense. Each prime factor can contribute at most one additional solution in the interval, plus the two trivial solutions. Therefore, the total number is ( 2 + k ), and since ( k leq log_2 m ), the bound is ( 2 + log_2 m ). So, putting it all together, the number of solutions ( x ) in the interval ([c, c + sqrt{m}]) such that ( x^2 equiv a pmod{m} ) is at most ( 2 + log_2 m ). I think that's the reasoning. It's a bit hand-wavy, but I believe the key idea is that the number of solutions is related to the number of prime factors, and within the interval, you can have at most one solution per prime factor plus two trivial ones, leading to the bound ( 2 + log_2 m ).</think>

🧠:<think>Alright, so I have this problem to consider: I need to look at all sets of ( n ) distinct positive integers where no three of them form an arithmetic progression. And then, among all these sets, I have to prove that there's one set which has the largest sum of the reciprocals of its elements. Hmm, okay.First, let me make sure I understand the problem correctly. An arithmetic progression is a sequence where each term after the first is obtained by adding a constant difference. So, if I have three numbers ( a ), ( b ), and ( c ), they form an arithmetic progression if ( b - a = c - b ), which simplifies to ( 2b = a + c ). So, in the context of this problem, we need to ensure that for any three elements in our set, this condition doesn't hold.Now, the goal is to find a set of ( n ) distinct positive integers with this property, and among all such sets, identify the one with the maximum sum of reciprocals. That is, we want to maximize ( sum_{k=1}^{n} frac{1}{a_k} ), where ( a_k ) are the elements of our set.Okay, so I need to think about how to approach this. Maybe I can start by considering smaller values of ( n ) and see if I can spot a pattern or a strategy that might generalize.Let's take ( n = 3 ) as a simple case. If ( n = 3 ), we need three distinct positive integers with no three-term arithmetic progression. But wait, if we have three numbers, they automatically form an arithmetic progression if they are equally spaced. So, to avoid that, we need to choose numbers that are not equally spaced.For example, instead of choosing 1, 2, 3, which form an arithmetic progression, we could choose 1, 2, 4. Let's check: ( 2 - 1 = 1 ) and ( 4 - 2 = 2 ), which are not equal, so that's good. Now, the sum of reciprocals would be ( frac{1}{1} + frac{1}{2} + frac{1}{4} = 1 + 0.5 + 0.25 = 1.75 ).Alternatively, if I choose 1, 3, 4, the reciprocals would be ( 1 + frac{1}{3} + frac{1}{4} approx 1 + 0.333 + 0.25 = 1.583 ), which is less than 1.75. So, in this case, 1, 2, 4 gives a higher sum of reciprocals.Wait, but is 1, 2, 4 the set with the maximum sum of reciprocals? Let me see another example. If I take 1, 2, 5, the sum is ( 1 + 0.5 + 0.2 = 1.7 ), which is less than 1.75. Similarly, 1, 2, 6 gives ( 1 + 0.5 + 0.166 approx 1.666 ), still less. So, it seems that 1, 2, 4 is indeed better.What if I choose higher numbers? For example, 2, 3, 5. The reciprocals sum to ( frac{1}{2} + frac{1}{3} + frac{1}{5} approx 0.5 + 0.333 + 0.2 = 1.033 ), which is much less. So, definitely, keeping the numbers as small as possible seems beneficial for maximizing the sum of reciprocals.But wait, in the case of ( n = 3 ), the set {1, 2, 4} is the one with the maximum sum of reciprocals. Is that always the case for larger ( n )?Let's try ( n = 4 ). We need four distinct positive integers with no three-term arithmetic progression. Let's try to build upon the previous set {1, 2, 4}. What number can we add next? If we add 5, we need to check if any three numbers form an arithmetic progression.Check subsets:- 1, 2, 3: Not in the set.- 1, 2, 4: Not an AP.- 1, 2, 5: ( 2 - 1 = 1 ), ( 5 - 2 = 3 ), not equal.- 1, 4, 5: ( 4 - 1 = 3 ), ( 5 - 4 = 1 ), not equal.- 2, 4, 5: ( 4 - 2 = 2 ), ( 5 - 4 = 1 ), not equal.So, {1, 2, 4, 5} is valid. The sum of reciprocals is ( 1 + 0.5 + 0.25 + 0.2 = 1.95 ).Alternatively, if I add 3 instead of 5 to the set {1, 2, 4}, we get {1, 2, 3, 4}. But in this set, 1, 2, 3 form an AP, and 2, 3, 4 form an AP, so that's not allowed.What if I add 6? Let's see: {1, 2, 4, 6}. Check for APs:- 1, 2, 3: Not in the set.- 1, 2, 4: Not an AP.- 1, 2, 6: ( 2 - 1 = 1 ), ( 6 - 2 = 4 ), not equal.- 1, 4, 6: ( 4 - 1 = 3 ), ( 6 - 4 = 2 ), not equal.- 2, 4, 6: ( 4 - 2 = 2 ), ( 6 - 4 = 2 ), so this is an AP! Oh, so {1, 2, 4, 6} is invalid because 2, 4, 6 form an AP.Therefore, adding 6 is not allowed. So, 5 seems to be the next number we can add. So, {1, 2, 4, 5} is a valid set with a higher sum of reciprocals.Is there another set with a higher sum? Let's try {1, 2, 3, 5}, but 1, 2, 3 is an AP, so that's invalid. What about {1, 3, 4, 5}? Let's check for APs:- 1, 3, 5: ( 3 - 1 = 2 ), ( 5 - 3 = 2 ), so that's an AP. So, invalid.How about {1, 2, 5, 7}? Let's check:- 1, 2, 3: Not in the set.- 1, 2, 5: Not an AP.- 1, 2, 7: Not an AP.- 1, 5, 7: Not an AP.- 2, 5, 7: ( 5 - 2 = 3 ), ( 7 - 5 = 2 ), not equal.So, {1, 2, 5, 7} is valid. The sum of reciprocals is ( 1 + 0.5 + 0.2 + 0.1428 approx 1.8428 ), which is less than 1.95. So, {1, 2, 4, 5} is better.Is there another set? Maybe {1, 3, 4, 6}. Check for APs:- 1, 3, 5: Not in the set.- 1, 3, 4: Not an AP.- 1, 3, 6: Not an AP.- 1, 4, 6: Not an AP.- 3, 4, 6: Not an AP.So, {1, 3, 4, 6} is valid. The sum of reciprocals is ( 1 + frac{1}{3} + frac{1}{4} + frac{1}{6} approx 1 + 0.333 + 0.25 + 0.166 approx 1.749 ), which is still less than 1.95.It seems that {1, 2, 4, 5} gives the highest sum of reciprocals for ( n = 4 ). So, maybe the strategy is to include the smallest possible numbers while avoiding arithmetic progressions.But wait, when ( n = 4 ), to include 5, we had to skip 3 and 6 because adding them would create an AP. So, the set {1, 2, 4, 5} is the minimal set in terms of numbers, but it's not the set with the smallest numbers because 3 is excluded.Is there a way to include 3 without creating an AP? Let's see. If we have {1, 2, 4, 3}, but then 1, 2, 3 is an AP, so that's invalid. Alternatively, if we arrange them differently, but any set containing 1, 2, 3 will have an AP. So, 3 cannot be included if we have 1 and 2.So, perhaps the optimal set is to include 1, 2, and then the next numbers that don't form an AP with any two previous numbers.Wait, that sounds like the greedy algorithm. Maybe the set is constructed by starting with 1, then adding the smallest possible number that doesn't form an AP with any two existing numbers.Let's try constructing such a set for ( n = 5 ). Starting with {1, 2, 4, 5}, what's the next number we can add?Check 6: Adding 6 would create an AP with 2 and 4? ( 2, 4, 6 ) is an AP, so 6 is invalid.Check 7: Adding 7. Check all pairs with existing numbers:- 1 and 2: 1, 2, 3 (not in set), 1, 2, 4 (already in set, but 1, 2, 7: not an AP).- 1 and 4: 1, 4, 7: ( 4 - 1 = 3 ), ( 7 - 4 = 3 ), so that's an AP! Oh, so adding 7 would create an AP with 1 and 4. So, 7 is invalid.Check 8: Adding 8. Check pairs:- 1 and 2: 1, 2, 3; 1, 2, 8: not an AP.- 1 and 4: 1, 4, 8: ( 4 - 1 = 3 ), ( 8 - 4 = 4 ), not equal.- 1 and 5: 1, 5, 9 (not in set).- 2 and 4: 2, 4, 6 (not in set).- 2 and 5: 2, 5, 8: ( 5 - 2 = 3 ), ( 8 - 5 = 3 ), so that's an AP! So, adding 8 would create an AP with 2 and 5. Invalid.Check 9: Adding 9. Check pairs:- 1 and 2: 1, 2, 3; 1, 2, 9: not an AP.- 1 and 4: 1, 4, 7 (not in set).- 1 and 5: 1, 5, 9: ( 5 - 1 = 4 ), ( 9 - 5 = 4 ), so that's an AP! So, adding 9 would create an AP with 1 and 5. Invalid.Check 10: Adding 10. Check pairs:- 1 and 2: 1, 2, 3; 1, 2, 10: not an AP.- 1 and 4: 1, 4, 7 (not in set).- 1 and 5: 1, 5, 9 (not in set).- 2 and 4: 2, 4, 6 (not in set).- 2 and 5: 2, 5, 8 (not in set).- 4 and 5: 4, 5, 6 (not in set).So, adding 10 seems okay. Check for any APs:- 1, 2, 10: Not an AP.- 1, 4, 10: Not an AP.- 1, 5, 10: Not an AP.- 2, 4, 10: ( 4 - 2 = 2 ), ( 10 - 4 = 6 ), not equal.- 2, 5, 10: ( 5 - 2 = 3 ), ( 10 - 5 = 5 ), not equal.- 4, 5, 10: Not an AP.So, {1, 2, 4, 5, 10} is a valid set. The sum of reciprocals is ( 1 + 0.5 + 0.25 + 0.2 + 0.1 = 2.05 ).Is there a better set? Let's see. Maybe if we choose different numbers. For example, {1, 2, 4, 8, 10}. Let's check:- 1, 2, 4: Not an AP.- 1, 2, 8: Not an AP.- 1, 2, 10: Not an AP.- 1, 4, 8: ( 4 - 1 = 3 ), ( 8 - 4 = 4 ), not equal.- 1, 4, 10: ( 4 - 1 = 3 ), ( 10 - 4 = 6 ), not equal.- 1, 8, 10: Not an AP.- 2, 4, 8: ( 4 - 2 = 2 ), ( 8 - 4 = 4 ), not equal.- 2, 4, 10: Not an AP.- 2, 8, 10: ( 8 - 2 = 6 ), ( 10 - 8 = 2 ), not equal.- 4, 8, 10: Not an AP.So, {1, 2, 4, 8, 10} is valid. The sum of reciprocals is ( 1 + 0.5 + 0.25 + 0.125 + 0.1 = 1.975 ), which is less than 2.05. So, {1, 2, 4, 5, 10} is better.Alternatively, what about {1, 2, 5, 7, 10}? Let's check:- 1, 2, 5: Not an AP.- 1, 2, 7: Not an AP.- 1, 2, 10: Not an AP.- 1, 5, 7: Not an AP.- 1, 5, 10: Not an AP.- 1, 7, 10: Not an AP.- 2, 5, 7: Not an AP.- 2, 5, 10: Not an AP.- 2, 7, 10: Not an AP.- 5, 7, 10: Not an AP.Sum of reciprocals: ( 1 + 0.5 + 0.2 + 0.1428 + 0.1 approx 1.9428 ), which is less than 2.05.So, it seems that {1, 2, 4, 5, 10} is the best so far.Wait, but when I added 5 to the set {1, 2, 4}, I had to skip 3 and 6, but adding 5 didn't create any APs. Then, adding 10 as the next number. Maybe there's a pattern here.I notice that the set {1, 2, 4, 5, 10} is constructed by taking the smallest possible numbers while avoiding any three-term AP. It seems like a greedy approach, where at each step, we add the smallest number that doesn't form an AP with any two already in the set.But is this always the optimal strategy? Let me think.Suppose I have a set S, and I add a new element x. If x is as small as possible, then 1/x is as large as possible, contributing more to the sum. So, to maximize the sum of reciprocals, we want the elements of the set to be as small as possible. However, we have the constraint that no three elements form an AP.Therefore, the optimal set should be the one where the elements are the smallest possible numbers that avoid forming any three-term APs. This seems to align with the greedy approach.But how do I formalize this? I need to prove that such a set exists and that it's unique in maximizing the sum of reciprocals.Wait, is it unique? For example, for ( n = 3 ), {1, 2, 4} is unique in the sense that any other set would have a larger number, leading to a smaller reciprocal sum. For ( n = 4 ), {1, 2, 4, 5} seems to be the unique set with the maximum sum. For ( n = 5 ), {1, 2, 4, 5, 10} is the set I found, but is there another set with the same sum or higher?Suppose I try {1, 2, 4, 7, 8}. Let's check for APs:- 1, 2, 4: Not an AP.- 1, 2, 7: Not an AP.- 1, 2, 8: Not an AP.- 1, 4, 7: ( 4 - 1 = 3 ), ( 7 - 4 = 3 ), so that's an AP! So, invalid.Okay, so that doesn't work. How about {1, 2, 5, 7, 8}?- 1, 2, 5: Not an AP.- 1, 2, 7: Not an AP.- 1, 2, 8: Not an AP.- 1, 5, 7: Not an AP.- 1, 5, 8: Not an AP.- 1, 7, 8: Not an AP.- 2, 5, 7: Not an AP.- 2, 5, 8: Not an AP.- 2, 7, 8: Not an AP.- 5, 7, 8: Not an AP.So, {1, 2, 5, 7, 8} is valid. Sum of reciprocals: ( 1 + 0.5 + 0.2 + 0.1428 + 0.125 approx 1.9678 ), which is still less than 2.05.So, {1, 2, 4, 5, 10} is better.Another attempt: {1, 2, 4, 7, 10}.- 1, 2, 4: Not an AP.- 1, 2, 7: Not an AP.- 1, 2, 10: Not an AP.- 1, 4, 7: AP! ( 4 - 1 = 3 ), ( 7 - 4 = 3 ). So, invalid.So, that doesn't work.Another idea: {1, 3, 4, 5, 10}.- 1, 3, 4: Not an AP.- 1, 3, 5: ( 3 - 1 = 2 ), ( 5 - 3 = 2 ), so that's an AP! Invalid.Hmm, tricky.Wait, maybe {1, 2, 4, 5, 10} is indeed the unique set with the maximum sum for ( n = 5 ).So, from these examples, it seems that constructing the set by choosing the smallest possible numbers at each step, avoiding the creation of any three-term AP, leads to the set with the maximum sum of reciprocals.But how do I generalize this to any ( n )? I need to show that such a set exists and that it's the one with the maximum sum.One approach could be to use induction. Assume that for a set of size ( k ), the set constructed by the greedy algorithm (choosing the smallest possible numbers without forming an AP) has the maximum sum of reciprocals. Then, show that adding the next smallest number that doesn't form an AP will maintain this property for ( k + 1 ).However, induction might be tricky here because the property of avoiding APs is not just about the new element but about all possible triples involving the new element and any two existing elements.Alternatively, maybe I can think in terms of partial orders or something like that.Wait, another idea: consider all possible sets of ( n ) distinct positive integers with no three-term AP. Among these sets, there must be one with the smallest possible maximum element, right? Because as numbers get larger, their reciprocals get smaller, so to maximize the sum, we want the numbers to be as small as possible.But actually, it's not just about the maximum element; it's about all the elements. So, the set with the smallest possible elements that avoid APs would have the largest sum of reciprocals.Therefore, the set that is lexicographically smallest in some sense, but ensuring no APs.Wait, maybe I can formalize this by considering the concept of minimal sets. If I can show that the set constructed by the greedy algorithm is minimal in some ordering, then it would have the maximum sum of reciprocals.Alternatively, think about the problem in terms of optimization. We want to maximize ( sum_{k=1}^{n} frac{1}{a_k} ), subject to the constraint that no three ( a_i, a_j, a_k ) form an AP.This is similar to an integer programming problem where we want to choose integers to maximize a function subject to certain constraints.But perhaps a more combinatorial approach is needed.Another thought: since reciprocals are decreasing functions, to maximize the sum, we need to have the elements as small as possible. So, the problem reduces to finding the smallest possible set (in terms of element size) that avoids three-term APs.Thus, the set with the smallest possible elements that satisfy the AP constraint will have the maximum sum of reciprocals.Therefore, I need to show that such a minimal set exists and is unique.Wait, but is it unique? From the examples above, for ( n = 3 ), {1, 2, 4} is the unique minimal set. For ( n = 4 ), {1, 2, 4, 5} is unique. For ( n = 5 ), {1, 2, 4, 5, 10} seems unique as well.So, perhaps for each ( n ), there's a unique minimal set that avoids three-term APs, and this set will have the maximum sum of reciprocals.But how do I prove this?Maybe I can use the concept of well-ordering. Since the set of positive integers is well-ordered, any non-empty set has a least element. So, starting from 1, we can build the set step by step, choosing the smallest possible number at each step that doesn't form an AP with any two existing numbers.This process will terminate after ( n ) steps, giving us a set of size ( n ) with no three-term APs. Moreover, since we always choose the smallest possible number, this set will have the smallest possible elements, leading to the maximum sum of reciprocals.But I need to make sure that this process doesn't get stuck before reaching ( n ) elements. That is, at each step, there exists a number that can be added without forming an AP.Is that always the case? For any ( n ), can we find such a set?Yes, because we can always choose numbers that are spaced out enough to avoid forming APs. For example, choosing powers of 2: {1, 2, 4, 8, 16, ...} ensures no three-term AP because the differences double each time, so no two differences can be equal. But this set might not be the minimal one in terms of element size, but it shows that such sets exist for any ( n ).However, the set constructed by the greedy algorithm is more minimal, as it includes smaller numbers where possible.So, putting it all together, the strategy is:1. Start with the smallest positive integer, 1.2. At each step, add the smallest integer that doesn't form a three-term AP with any two already in the set.3. Continue until the set has ( n ) elements.This set will have the smallest possible elements, thus maximizing the sum of reciprocals, and it will avoid any three-term APs by construction.Therefore, such a set exists and is unique, and it has the largest sum of reciprocals among all sets of ( n ) distinct positive integers with no three-term AP.I think this approach covers the necessary steps to prove the statement. It uses the greedy algorithm to construct the desired set and leverages the properties of reciprocals and arithmetic progressions to ensure optimality.</think>

🧠:<think>Alright, so I have this problem about a function ( g(x) ) that's defined between ( x = -5 ) and ( x = 5 ). The graph of ( g(x) ) is made up of three different segments: 1. A downward-opening parabola from ( x = -5 ) to ( x = -1 ).2. A straight line from ( x = -1 ) to ( x = 3 ).3. An upward-opening parabola from ( x = 3 ) to ( x = 5 ).And the question is asking how many values of ( x ) satisfy the equation ( g(g(x)) = 3 ).Hmm, okay. So, I need to figure out how many solutions exist for ( g(g(x)) = 3 ). To do this, I think I need to approach it step by step. Maybe I should first find all the values of ( x ) such that ( g(x) = 3 ), and then for each of those values, find how many ( x ) satisfy ( g(x) = ) that value. So, essentially, it's like solving ( g(x) = a ) where ( a ) are the solutions to ( g(a) = 3 ).Let me write this down more formally. Let’s denote:1. First, find all ( x ) such that ( g(x) = 3 ). Let's call these solutions ( x_1, x_2, ldots, x_n ).2. Then, for each ( x_i ), find all ( y ) such that ( g(y) = x_i ). The total number of such ( y ) will give the total number of solutions to ( g(g(y)) = 3 ).Okay, that makes sense. So, I need to break it down into two parts: first solving ( g(x) = 3 ), and then for each solution, solving ( g(y) = x_i ).But, before I can do that, I need to know the exact form of ( g(x) ) in each interval. The problem describes the graph, but it doesn't give the equations. So, I might need to assume some equations based on the description.Let me think about how to model each segment.1. First segment: Downward-opening parabola from ( x = -5 ) to ( x = -1 ). - A downward-opening parabola has the general form ( g(x) = -a(x - h)^2 + k ), where ( (h, k) ) is the vertex. - Since it's a downward-opening parabola, the coefficient is negative. - I don't know the vertex or the scaling factor, so I might need to make some assumptions or see if I can get more information.2. Second segment: Straight line from ( x = -1 ) to ( x = 3 ). - A straight line can be written as ( g(x) = mx + b ). - Again, without specific points, it's hard to determine the exact equation, but maybe I can figure it out based on continuity or other information.3. Third segment: Upward-opening parabola from ( x = 3 ) to ( x = 5 ). - An upward-opening parabola has the general form ( g(x) = a(x - h)^2 + k ), where ( (h, k) ) is the vertex. - Since it's upward-opening, the coefficient is positive.Hmm, maybe I can assume that the function is continuous at the joining points ( x = -1 ) and ( x = 3 ). So, the value of the first parabola at ( x = -1 ) should be equal to the value of the straight line at ( x = -1 ), and similarly, the straight line at ( x = 3 ) should equal the upward-opening parabola at ( x = 3 ).But without specific values, it's still a bit challenging. Maybe I can assign some specific forms to these segments for simplicity.Let me try to define each segment with some equations:1. First segment (downward-opening parabola from ( x = -5 ) to ( x = -1 )): - Let's assume the vertex is at ( x = -3 ) since it's midway between ( -5 ) and ( -1 ). So, the equation would be ( g(x) = -a(x + 3)^2 + k ). - Let's choose ( a = 1 ) for simplicity, so ( g(x) = -(x + 3)^2 + k ). - To find ( k ), maybe we can set the value at ( x = -5 ) or ( x = -1 ). Let's say at ( x = -5 ), ( g(-5) = -(-5 + 3)^2 + k = -(-2)^2 + k = -4 + k ). Similarly, at ( x = -1 ), ( g(-1) = -(-1 + 3)^2 + k = -(2)^2 + k = -4 + k ). - Wait, that would mean the parabola starts and ends at the same value, which is ( -4 + k ). But since it's a downward-opening parabola, maybe the vertex is the maximum point. So, the maximum value is ( k ), which occurs at ( x = -3 ). - Let me choose ( k = 4 ) so that the maximum is 4. So, the equation becomes ( g(x) = -(x + 3)^2 + 4 ). - Therefore, at ( x = -5 ) and ( x = -1 ), ( g(x) = -4 + 4 = 0 ). So, the parabola goes from ( (-5, 0) ) up to ( (-3, 4) ) and back down to ( (-1, 0) ).2. Second segment (straight line from ( x = -1 ) to ( x = 3 )): - We know that at ( x = -1 ), ( g(-1) = 0 ) from the first segment. - Let's assume that at ( x = 3 ), the value should match the third segment. Let's say the straight line goes from ( (-1, 0) ) to ( (3, c) ), where ( c ) is some value. - But we need more information to find the slope. Maybe we can assume it's a linear increase or decrease. Let's suppose it's a straight line with a slope of 1 for simplicity. - So, the equation would be ( g(x) = x + b ). Plugging in ( x = -1 ), ( g(-1) = -1 + b = 0 ), so ( b = 1 ). - Therefore, the equation is ( g(x) = x + 1 ) for ( -1 < x leq 3 ). - At ( x = 3 ), ( g(3) = 3 + 1 = 4 ).3. Third segment (upward-opening parabola from ( x = 3 ) to ( x = 5 )): - At ( x = 3 ), the value is 4, so the parabola starts at ( (3, 4) ). - Let's assume the vertex is at ( x = 4 ) since it's midway between 3 and 5. - The general form is ( g(x) = a(x - 4)^2 + k ). - Since it's upward-opening, ( a > 0 ). Let's choose ( a = 1 ) for simplicity. - So, ( g(x) = (x - 4)^2 + k ). - At ( x = 3 ), ( g(3) = (3 - 4)^2 + k = 1 + k = 4 ). Therefore, ( k = 3 ). - So, the equation is ( g(x) = (x - 4)^2 + 3 ). - At ( x = 5 ), ( g(5) = (5 - 4)^2 + 3 = 1 + 3 = 4 ).Alright, so now I have defined ( g(x) ) in each interval:1. For ( -5 leq x leq -1 ), ( g(x) = -(x + 3)^2 + 4 ).2. For ( -1 < x leq 3 ), ( g(x) = x + 1 ).3. For ( 3 < x leq 5 ), ( g(x) = (x - 4)^2 + 3 ).Now, I need to find all ( x ) such that ( g(g(x)) = 3 ). As I thought earlier, I need to first find all ( x ) such that ( g(x) = 3 ), and then for each of those solutions, find all ( y ) such that ( g(y) = x_i ).So, let's start by solving ( g(x) = 3 ).Step 1: Solve ( g(x) = 3 )We'll consider each interval separately.1. For ( -5 leq x leq -1 ): ( g(x) = -(x + 3)^2 + 4 = 3 ) Let's solve this: [ -(x + 3)^2 + 4 = 3 -(x + 3)^2 = -1 (x + 3)^2 = 1 x + 3 = pm 1 x = -3 pm 1 ] So, ( x = -3 + 1 = -2 ) or ( x = -3 - 1 = -4 ). Both ( x = -2 ) and ( x = -4 ) are within ( -5 leq x leq -1 ), so these are valid solutions.2. For ( -1 < x leq 3 ): ( g(x) = x + 1 = 3 ) Solving: [ x + 1 = 3 x = 2 ] ( x = 2 ) is within ( -1 < x leq 3 ), so this is a valid solution.3. For ( 3 < x leq 5 ): ( g(x) = (x - 4)^2 + 3 = 3 ) Solving: [ (x - 4)^2 + 3 = 3 (x - 4)^2 = 0 x - 4 = 0 x = 4 ] ( x = 4 ) is within ( 3 < x leq 5 ), so this is a valid solution.So, the solutions to ( g(x) = 3 ) are ( x = -4, -2, 2, 4 ).Step 2: For each ( x_i ) where ( g(x_i) = 3 ), find all ( y ) such that ( g(y) = x_i )So, now, for each ( x_i ) in ( { -4, -2, 2, 4 } ), we need to solve ( g(y) = x_i ).Let's do this one by one.1. Case 1: Solve ( g(y) = -4 ) We need to find all ( y ) such that ( g(y) = -4 ). Let's check each interval: a. For ( -5 leq y leq -1 ): ( g(y) = -(y + 3)^2 + 4 = -4 ) Solving: [ -(y + 3)^2 + 4 = -4 -(y + 3)^2 = -8 (y + 3)^2 = 8 y + 3 = pm sqrt{8} y = -3 pm 2sqrt{2} ] Calculating approximate values: ( sqrt{8} approx 2.828 ), so: ( y approx -3 + 2.828 approx -0.172 ) and ( y approx -3 - 2.828 approx -5.828 ). Now, check if these are within ( -5 leq y leq -1 ): - ( y approx -0.172 ) is not within this interval. - ( y approx -5.828 ) is less than ( -5 ), so it's not within this interval either. So, no solutions in this interval. b. For ( -1 < y leq 3 ): ( g(y) = y + 1 = -4 ) Solving: [ y + 1 = -4 y = -5 ] But ( y = -5 ) is not in ( -1 < y leq 3 ), so no solution here. c. For ( 3 < y leq 5 ): ( g(y) = (y - 4)^2 + 3 = -4 ) Solving: [ (y - 4)^2 + 3 = -4 (y - 4)^2 = -7 ] Since the square of a real number can't be negative, there are no real solutions here. So, there are no solutions for ( g(y) = -4 ).2. Case 2: Solve ( g(y) = -2 ) Now, find all ( y ) such that ( g(y) = -2 ). Again, check each interval: a. For ( -5 leq y leq -1 ): ( g(y) = -(y + 3)^2 + 4 = -2 ) Solving: [ -(y + 3)^2 + 4 = -2 -(y + 3)^2 = -6 (y + 3)^2 = 6 y + 3 = pm sqrt{6} y = -3 pm sqrt{6} ] Calculating approximate values: ( sqrt{6} approx 2.449 ), so: ( y approx -3 + 2.449 approx -0.551 ) and ( y approx -3 - 2.449 approx -5.449 ). Checking intervals: - ( y approx -0.551 ) is not in ( -5 leq y leq -1 ). - ( y approx -5.449 ) is less than ( -5 ), so not in this interval. So, no solutions here. b. For ( -1 < y leq 3 ): ( g(y) = y + 1 = -2 ) Solving: [ y + 1 = -2 y = -3 ] But ( y = -3 ) is not in ( -1 < y leq 3 ), so no solution. c. For ( 3 < y leq 5 ): ( g(y) = (y - 4)^2 + 3 = -2 ) Solving: [ (y - 4)^2 + 3 = -2 (y - 4)^2 = -5 ] Again, no real solutions since the square can't be negative. So, no solutions for ( g(y) = -2 ).3. Case 3: Solve ( g(y) = 2 ) Find all ( y ) such that ( g(y) = 2 ). Checking each interval: a. For ( -5 leq y leq -1 ): ( g(y) = -(y + 3)^2 + 4 = 2 ) Solving: [ -(y + 3)^2 + 4 = 2 -(y + 3)^2 = -2 (y + 3)^2 = 2 y + 3 = pm sqrt{2} y = -3 pm sqrt{2} ] Approximate values: ( sqrt{2} approx 1.414 ), so: ( y approx -3 + 1.414 approx -1.586 ) and ( y approx -3 - 1.414 approx -4.414 ). Checking intervals: - ( y approx -1.586 ) is between ( -5 ) and ( -1 ), so it's valid. - ( y approx -4.414 ) is also within ( -5 leq y leq -1 ), so valid. So, two solutions here: ( y = -3 + sqrt{2} ) and ( y = -3 - sqrt{2} ). b. For ( -1 < y leq 3 ): ( g(y) = y + 1 = 2 ) Solving: [ y + 1 = 2 y = 1 ] ( y = 1 ) is within ( -1 < y leq 3 ), so that's another solution. c. For ( 3 < y leq 5 ): ( g(y) = (y - 4)^2 + 3 = 2 ) Solving: [ (y - 4)^2 + 3 = 2 (y - 4)^2 = -1 ] No real solutions here. So, total solutions for ( g(y) = 2 ) are ( y = -3 + sqrt{2} ), ( y = -3 - sqrt{2} ), and ( y = 1 ). That's three solutions.4. Case 4: Solve ( g(y) = 4 ) Now, find all ( y ) such that ( g(y) = 4 ). Checking each interval: a. For ( -5 leq y leq -1 ): ( g(y) = -(y + 3)^2 + 4 = 4 ) Solving: [ -(y + 3)^2 + 4 = 4 -(y + 3)^2 = 0 (y + 3)^2 = 0 y + 3 = 0 y = -3 ] ( y = -3 ) is within ( -5 leq y leq -1 ), so that's a solution. b. For ( -1 < y leq 3 ): ( g(y) = y + 1 = 4 ) Solving: [ y + 1 = 4 y = 3 ] ( y = 3 ) is within ( -1 < y leq 3 ), so that's another solution. c. For ( 3 < y leq 5 ): ( g(y) = (y - 4)^2 + 3 = 4 ) Solving: [ (y - 4)^2 + 3 = 4 (y - 4)^2 = 1 y - 4 = pm 1 y = 4 pm 1 y = 5 text{ or } y = 3 ] Checking intervals: - ( y = 5 ) is within ( 3 < y leq 5 ), so valid. - ( y = 3 ) is at the boundary. Since this interval is ( 3 < y leq 5 ), ( y = 3 ) is not included here, but it was already considered in the previous interval. So, ( y = 5 ) is another solution. So, total solutions for ( g(y) = 4 ) are ( y = -3 ), ( y = 3 ), and ( y = 5 ). That's three solutions.Step 3: Compile All SolutionsNow, let's compile all the solutions we found:- For ( g(y) = -4 ): No solutions.- For ( g(y) = -2 ): No solutions.- For ( g(y) = 2 ): ( y = -3 + sqrt{2} ), ( y = -3 - sqrt{2} ), ( y = 1 ).- For ( g(y) = 4 ): ( y = -3 ), ( y = 3 ), ( y = 5 ).So, in total, we have ( 3 + 3 = 6 ) solutions.Wait a minute, but let me double-check if all these solutions are valid and within their respective intervals.- ( y = -3 + sqrt{2} approx -1.586 ): Within ( -5 leq y leq -1 ). Wait, ( -1.586 ) is actually between ( -5 ) and ( -1 ), so it's valid.- ( y = -3 - sqrt{2} approx -4.414 ): Within ( -5 leq y leq -1 ), valid.- ( y = 1 ): Within ( -1 < y leq 3 ), valid.- ( y = -3 ): Within ( -5 leq y leq -1 ), valid.- ( y = 3 ): Within ( -1 < y leq 3 ), valid.- ( y = 5 ): Within ( 3 < y leq 5 ), valid.So, all six solutions are valid.But hold on, let me check if any of these ( y ) values when plugged back into ( g(g(y)) ) actually result in 3.Wait, we already ensured that ( g(y) = x_i ) where ( g(x_i) = 3 ), so by transitivity, ( g(g(y)) = g(x_i) = 3 ). So, all these solutions should be valid.But just to make sure, let's pick one example and verify.Let's take ( y = -3 + sqrt{2} approx -1.586 ). Then, ( g(y) = 2 ), and ( g(2) = 3 ). So, ( g(g(y)) = 3 ). That's correct.Another example: ( y = 5 ). Then, ( g(5) = 4 ), and ( g(4) = 3 ). So, ( g(g(5)) = 3 ). Correct.And ( y = 3 ). ( g(3) = 4 ), and ( g(4) = 3 ). So, ( g(g(3)) = 3 ). Correct.Okay, seems like all these solutions are valid.So, in total, we have six solutions: ( y = -3 - sqrt{2} ), ( y = -3 + sqrt{2} ), ( y = 1 ), ( y = -3 ), ( y = 3 ), and ( y = 5 ).Therefore, the number of values of ( x ) that satisfy ( g(g(x)) = 3 ) is 6.But wait, hold on. The initial solutions for ( g(x) = 3 ) were ( x = -4, -2, 2, 4 ). Then, for each of these, we found the number of ( y ) such that ( g(y) = x_i ). For ( x_i = -4 ): 0 solutions.For ( x_i = -2 ): 0 solutions.For ( x_i = 2 ): 3 solutions.For ( x_i = 4 ): 3 solutions.So, total solutions: 0 + 0 + 3 + 3 = 6.But let me think again about the first case where ( x_i = -4 ). We found no solutions for ( g(y) = -4 ). Similarly, for ( x_i = -2 ), no solutions. So, only the cases where ( x_i = 2 ) and ( x_i = 4 ) contribute solutions.So, the total number of solutions is indeed 6.But wait, in the original problem statement, it was mentioned that the function is defined only for ( x ) between ( -5 ) and ( 5 ) inclusive. So, all the solutions we found are within this domain, so we don't have to worry about any extraneous solutions.Is there any chance I might have missed some solutions? Let me quickly recap:- Solved ( g(x) = 3 ): 4 solutions.- For each of these 4, solved ( g(y) = x_i ): - ( x_i = -4 ): 0 - ( x_i = -2 ): 0 - ( x_i = 2 ): 3 - ( x_i = 4 ): 3- Total: 6 solutions.Yes, that seems consistent.So, my final answer is that there are 6 values of ( x ) that satisfy ( g(g(x)) = 3 ).But wait, hold on. Let me double-check the equations I assumed for each segment because if I made a mistake in defining ( g(x) ), it could affect the solutions.Original definitions:1. ( -5 leq x leq -1 ): ( g(x) = -(x + 3)^2 + 4 )2. ( -1 < x leq 3 ): ( g(x) = x + 1 )3. ( 3 < x leq 5 ): ( g(x) = (x - 4)^2 + 3 )I think these are consistent with the given graph description. The first segment is a downward-opening parabola with vertex at ( (-3, 4) ), goes through ( (-5, 0) ) and ( (-1, 0) ). The second segment is a straight line from ( (-1, 0) ) to ( (3, 4) ), which with slope 1 makes sense because ( (4 - 0)/(3 - (-1)) = 1 ). The third segment is an upward-opening parabola starting at ( (3, 4) ) and going to ( (5, 4) ) with vertex at ( (4, 3) ). So, yes, these equations seem correct.Therefore, I think my solutions are correct.Final AnswerThe number of values of ( x ) that satisfy ( g(g(x)) = 3 ) is (boxed{6}).</think>

🧠:<think>Alright, so I have this problem where I need to prove that for two distinct positive integers ( m ) and ( n ), there exists a real number ( x ) such that both fractional parts ( {xn} ) and ( {xm} ) are between ( frac{1}{3} ) and ( frac{2}{3} ). Hmm, fractional parts, okay. I remember that the fractional part of a number ( y ) is just ( y ) minus the integer part of ( y ), so it's like the decimal part, right? For example, ( {3.14} = 0.14 ).So, I need to find an ( x ) such that when I multiply it by ( m ) and ( n ), the decimal parts of both results are at least ( frac{1}{3} ) and at most ( frac{2}{3} ). That means neither ( {xm} ) nor ( {xn} ) should be too close to 0 or 1. They should be in the middle third of the interval [0,1).I think this has something to do with the distribution of multiples of ( m ) and ( n ) modulo 1. Maybe something like the pigeonhole principle? Or perhaps using some properties of irrational numbers? Wait, but ( m ) and ( n ) are integers, so ( xm ) and ( xn ) would be real numbers, but their fractional parts depend on ( x ).Let me think. If I can find an ( x ) such that both ( xm ) and ( xn ) have their fractional parts in [1/3, 2/3], that would solve the problem. Maybe I can construct such an ( x ) by considering multiples of some number.Wait, what if I consider ( x ) as a rational number? Let's say ( x = frac{p}{q} ) where ( p ) and ( q ) are integers. Then ( xm = frac{pm}{q} ) and ( xn = frac{pn}{q} ). The fractional parts would depend on ( pm ) modulo ( q ) and ( pn ) modulo ( q ). Maybe if I choose ( q ) appropriately, I can make sure that both fractional parts fall into the desired interval.Alternatively, perhaps I can use the concept of simultaneous Diophantine approximations. Since ( m ) and ( n ) are integers, maybe I can find an ( x ) such that both ( xm ) and ( xn ) are close to integers plus 1/3 or 2/3. Hmm, not sure.Wait, another idea: if I can find an integer ( k ) such that ( frac{k}{m} ) and ( frac{k}{n} ) have fractional parts in [1/3, 2/3], then setting ( x = frac{k}{mn} ) might work. Let me see. If ( x = frac{k}{mn} ), then ( xm = frac{k}{n} ) and ( xn = frac{k}{m} ). So, the fractional parts of ( frac{k}{n} ) and ( frac{k}{m} ) should be in [1/3, 2/3].But I need to ensure that such a ( k ) exists. Maybe by the Chinese Remainder Theorem or something? Let me think about residues modulo 3n or something.Wait, maybe I should think about distributing multiples of ( m ) and ( n ) modulo 1. Since ( m ) and ( n ) are integers, ( xm ) and ( xn ) are just scaling ( x ). So, by adjusting ( x ), I can make their fractional parts fall anywhere in [0,1). But the challenge is to make both of them fall into [1/3, 2/3] simultaneously.Perhaps using the pigeonhole principle on intervals. If I partition [0,1) into three intervals: [0,1/3), [1/3,2/3], and [2/3,1). Then, for each ( x ), ( xm ) and ( xn ) will each fall into one of these three intervals. So, there are 3 choices for ( xm ) and 3 choices for ( xn ), making 9 possible combinations.But I want both ( xm ) and ( xn ) to be in the middle interval [1/3,2/3]. So, is it guaranteed that such an ( x ) exists?Wait, actually, maybe not directly. Because as ( x ) changes, the positions of ( xm ) and ( xn ) modulo 1 change in a correlated way. They aren't independent.Hmm, perhaps I need to consider the joint distribution of ( xm ) and ( xn ) modulo 1. Since ( m ) and ( n ) are integers, ( xm ) and ( xn ) can be seen as points on a torus ( mathbb{T}^2 ), and the trajectory as ( x ) varies is a straight line with slope ( m/n ) or something like that.But maybe this is getting too abstract. Let me try to think of it in terms of equations. I need:( frac{1}{3} leq {xm} leq frac{2}{3} ) ( frac{1}{3} leq {xn} leq frac{2}{3} )Which means:( xm = a + alpha ) ( xn = b + beta )where ( a ) and ( b ) are integers, and ( alpha, beta ) are in [1/3, 2/3].So, ( x = frac{a + alpha}{m} = frac{b + beta}{n} ). Therefore, we have:( frac{a + alpha}{m} = frac{b + beta}{n} )Which implies:( n(a + alpha) = m(b + beta) )Or:( na - mb = mbeta - nalpha )Now, ( na - mb ) is an integer because ( a ) and ( b ) are integers. On the right side, ( mbeta - nalpha ) must also be an integer. Since ( alpha, beta in [1/3, 2/3] ), ( mbeta ) and ( nalpha ) are real numbers. So, their difference must be an integer.But ( mbeta - nalpha ) is in the range:Minimum: ( m*(1/3) - n*(2/3) = frac{m - 2n}{3} ) Maximum: ( m*(2/3) - n*(1/3) = frac{2m - n}{3} )So, the integer ( na - mb ) must lie in the interval ( [frac{m - 2n}{3}, frac{2m - n}{3}] ). Therefore, there exists an integer ( k ) such that:( frac{m - 2n}{3} leq k leq frac{2m - n}{3} )Which can be rewritten as:( m - 2n leq 3k leq 2m - n )So, ( 3k ) must be between ( m - 2n ) and ( 2m - n ). Since ( m ) and ( n ) are positive integers, and distinct, this interval must contain at least one integer multiple of 3.Wait, is this always true? Let's see. The length of the interval is ( (2m - n) - (m - 2n) = m + n ). Since ( m ) and ( n ) are positive, the interval has a positive length.But does this interval necessarily contain a multiple of 3? Not necessarily. For example, if ( m = 1 ) and ( n = 2 ), then the interval is ( 1 - 4 = -3 ) to ( 2 - 2 = 0 ), so from -3 to 0. That does contain 0, which is a multiple of 3. Hmm, okay.Another example: ( m = 2 ), ( n = 1 ). The interval is ( 2 - 2 = 0 ) to ( 4 - 1 = 3 ). So, from 0 to 3. That contains 0, 3, which are multiples of 3.Wait, let's try ( m = 4 ), ( n = 1 ). Then the interval is ( 4 - 2 = 2 ) to ( 8 - 1 = 7 ). So, 2 to 7. The multiples of 3 in this interval are 3 and 6. So yes, there are multiples.Wait, another one: ( m = 3 ), ( n = 1 ). The interval is ( 3 - 2 = 1 ) to ( 6 - 1 = 5 ). So, 1 to 5. The multiples of 3 here are 3. So, yes, 3 is there.Wait, what about ( m = 5 ), ( n = 2 ). The interval is ( 5 - 4 = 1 ) to ( 10 - 2 = 8 ). So, 1 to 8. The multiples of 3 here are 3, 6. So, exists.Wait, is there a case where the interval doesn't contain a multiple of 3? Let's see. Suppose ( m = 1 ), ( n = 1 ). But they have to be distinct, so ( m = 1 ), ( n = 2 ) as before. The interval is -3 to 0, which includes 0, which is multiple of 3.Wait, maybe if ( m ) and ( n ) are such that the interval is between two non-multiples of 3, but with length greater than 3, then it must contain a multiple. But actually, the interval's length is ( m + n ), which is at least 1 + 2 = 3? Wait, ( m ) and ( n ) are distinct positive integers, so the smallest possible ( m + n ) is 1 + 2 = 3.So, if the interval has length at least 3, it must contain at least one multiple of 3. Wait, is that true? For example, the interval from 1 to 4 has length 3, and contains 3. From 2 to 5, length 3, contains 3. From 3 to 6, length 3, contains 3, 6. So, yes, it seems that any interval of length 3 on the real line must contain at least one multiple of 3.But wait, actually, no. For example, the interval from 0.1 to 3.1 has length 3, and contains 3, which is a multiple. But what about the interval from 0.5 to 3.5? It contains 3. So, still contains a multiple. Wait, but if the interval is entirely within, say, 1 to 4, it still contains 3. So, actually, any interval of length 3 must contain at least one multiple of 3, because the distance between multiples of 3 is 3.Therefore, the interval ( [frac{m - 2n}{3}, frac{2m - n}{3}] ) has length ( frac{m + n}{3} ), which, since ( m ) and ( n ) are positive integers, is at least 1. But to guarantee containing a multiple of 3, the length needs to be at least 3? Hmm, wait, no.Wait, actually, the length is ( frac{m + n}{3} ). Since ( m ) and ( n ) are at least 1, the length is at least ( frac{1 + 2}{3} = 1 ). So, it's not necessarily of length 3. So, my previous reasoning doesn't hold.Wait, so perhaps the interval might not contain a multiple of 3 if the length is less than 3. For example, suppose ( m = 1 ), ( n = 1 ) (but they have to be distinct, so maybe ( m = 1 ), ( n = 2 )). Then the interval is from ( frac{1 - 4}{3} = -1 ) to ( frac{2 - 2}{3} = 0 ). So, from -1 to 0, which includes 0, a multiple.Another example: ( m = 2 ), ( n = 3 ). Then the interval is ( frac{2 - 6}{3} = -frac{4}{3} ) to ( frac{4 - 3}{3} = frac{1}{3} ). So, from -1.333 to 0.333. This interval includes 0, which is a multiple.Wait, another case: ( m = 4 ), ( n = 5 ). Then the interval is ( frac{4 - 10}{3} = -2 ) to ( frac{8 - 5}{3} = 1 ). So, from -2 to 1, which includes -3, 0. So, multiple of 3 exists.Wait, maybe it's always the case that the interval contains 0 or another multiple because of the negative side. So, perhaps even if the positive side is short, the negative side includes a multiple.Wait, but if ( m ) and ( n ) are such that the lower bound is greater than -3, but the upper bound is less than 0, would that include a multiple? For example, if ( m = 1 ), ( n = 1 ), but they have to be distinct, so ( m =1 ), ( n = 2 ), as before, which gives an interval from -1 to 0, which includes 0.Wait, is there any case where the interval doesn't include a multiple of 3? Maybe if the interval is entirely within (0,1), but then since it's length is at least 1, it might not contain a multiple of 3. Wait, for example, suppose ( m = 3 ), ( n = 3 ), but they have to be distinct. So, ( m = 3 ), ( n = 4 ). Then the interval is ( frac{3 - 8}{3} = -frac{5}{3} ) to ( frac{6 - 4}{3} = frac{2}{3} ). So, from -1.666 to 0.666. This includes 0, which is a multiple.Wait, maybe it's always the case that the interval will include 0 or another multiple because of the subtraction. Hmm, maybe I'm overcomplicating.Alternatively, perhaps I should approach this problem using the concept of Beatty sequences or something similar, but I'm not sure.Wait, another thought: if I can find an integer ( k ) such that ( k ) is congruent to something modulo ( 3m ) and ( 3n ), then setting ( x = frac{k}{3mn} ) or something like that might work.Wait, let me try to use the Chinese Remainder Theorem. If ( m ) and ( n ) are coprime, then there exists a solution to the congruence equations:( k equiv a pmod{3m} ) ( k equiv b pmod{3n} )for any ( a ) and ( b ). But I need to find ( k ) such that ( { frac{k}{m} } ) and ( { frac{k}{n} } ) are in [1/3, 2/3]. Hmm, not sure.Wait, another idea: Let me consider the fractional parts as points on a circle, or the unit interval [0,1). If I can cover all possible combinations of fractional parts for ( xm ) and ( xn ), then by continuity, there must be some ( x ) where both are in the middle third.Wait, actually, since ( x ) can be any real number, as ( x ) increases, ( xm ) and ( xn ) increase continuously, and their fractional parts cycle through [0,1). So, perhaps by the intermediate value theorem, there must be some ( x ) where both fractional parts are in [1/3, 2/3].But I need to formalize this idea.Let me consider the function ( f(x) = ({xm}, {xn}) ). As ( x ) increases from 0 to 1, ( f(x) ) moves from (0,0) to (m mod 1, n mod 1), but since ( m ) and ( n ) are integers, actually ( f(x) ) cycles through the unit square [0,1) x [0,1) as ( x ) increases.But since ( m ) and ( n ) are distinct, the trajectory of ( f(x) ) is a straight line with slope ( m/n ) (assuming ( n neq 0 )), wrapping around the torus.Since ( m ) and ( n ) are integers, if they are coprime, the line will densely cover the torus. But even if they are not coprime, the line will cover a subgrid.So, the image of ( f(x) ) is dense in some subgrid of [0,1)^2. Therefore, it must come arbitrarily close to any point, including the region [1/3, 2/3] x [1/3, 2/3].But does it actually enter that region?Hmm, since the movement is continuous, and the region is open, if the trajectory passes through the region, then yes. But how do I ensure that it does?Wait, perhaps by considering that the trajectory wraps around, and given the periodicity, it must pass through every possible combination, including the middle third in both coordinates.Alternatively, maybe I can use the fact that the set of ( x ) where ( {xm} in [1/3, 2/3] ) has measure 1/3, and similarly for ( xn ). Since these are independent events in some sense, the intersection should have positive measure, so such ( x ) exists.But I'm not sure about the independence.Wait, actually, in ergodic theory, for irrational rotations, the fractional parts are uniformly distributed. But here, ( m ) and ( n ) are integers, so the rotation is rational, and the trajectory is periodic. So, the distribution is not uniform, but it's dense in a lattice.Wait, maybe I can use the pigeonhole principle on the periods. Let's say the period is ( L = text{lcm}(m, n) ). Then, over the interval ( x in [0, L] ), the fractional parts ( {xm} ) and ( {xn} ) will cycle through their periods.But I'm not sure how to apply pigeonhole here.Wait, going back to my initial idea, if I can find an integer ( k ) such that ( k equiv a pmod{3m} ) and ( k equiv b pmod{3n} ) with ( a ) and ( b ) chosen such that ( { frac{k}{m} } ) and ( { frac{k}{n} } ) are in [1/3, 2/3], then ( x = frac{k}{mn} ) would work.But how to choose ( a ) and ( b )? Let me think.If I set ( a = m times 1/3 ) and ( b = n times 1/3 ), but ( a ) and ( b ) need to be integers. Hmm, not directly.Wait, perhaps instead, set ( k = 3t + r ), where ( r ) is chosen such that ( { frac{k}{m} } ) and ( { frac{k}{n} } ) are in [1/3, 2/3].But I'm not sure.Wait, another approach: Let me consider ( x ) such that ( xm = a + alpha ) and ( xn = b + beta ), where ( a, b ) are integers and ( alpha, beta in [1/3, 2/3] ). Then, ( x = frac{a + alpha}{m} = frac{b + beta}{n} ), so ( n(a + alpha) = m(b + beta) ).This implies ( na - mb = mbeta - nalpha ). The left side is an integer, so the right side must also be an integer. Let me denote ( k = na - mb ), which is an integer. Therefore, ( k = mbeta - nalpha ).Since ( alpha, beta in [1/3, 2/3] ), ( mbeta in [m/3, 2m/3] ) and ( nalpha in [n/3, 2n/3] ). Therefore, ( mbeta - nalpha in [m/3 - 2n/3, 2m/3 - n/3] ).So, ( k ) must lie in the interval ( [frac{m - 2n}{3}, frac{2m - n}{3}] ). Since ( k ) is an integer, this interval must contain at least one integer.But is this interval guaranteed to contain an integer? Let's see.The length of the interval is ( frac{2m - n}{3} - frac{m - 2n}{3} = frac{m + n}{3} ). Since ( m ) and ( n ) are positive integers, ( m + n geq 3 ) (since they are distinct, minimal case ( m=1, n=2 )). Therefore, the length is at least 1.By the pigeonhole principle, an interval of length 1 on the real line must contain at least one integer. Therefore, the interval ( [frac{m - 2n}{3}, frac{2m - n}{3}] ) must contain at least one integer ( k ).Therefore, such an integer ( k ) exists. Once ( k ) is found, we can solve for ( a ) and ( b ):( na - mb = k )Which is a linear Diophantine equation. Since ( gcd(m, n) ) divides ( k ), and ( k ) is an integer, there exist integers ( a ) and ( b ) satisfying this equation.Once ( a ) and ( b ) are found, we can set ( x = frac{a + alpha}{m} ), where ( alpha ) is chosen such that ( alpha = frac{k + mb}{n} ). Wait, maybe I need to express ( alpha ) in terms of ( k ), ( m ), ( n ), ( a ), and ( b ).From ( k = mbeta - nalpha ), we have ( alpha = frac{mbeta - k}{n} ). Since ( beta in [1/3, 2/3] ), ( mbeta in [m/3, 2m/3] ), so ( alpha = frac{mbeta - k}{n} ) must be in [1/3, 2/3].Therefore, we can choose ( beta ) such that ( alpha ) falls into [1/3, 2/3]. Since ( beta ) can be adjusted within [1/3, 2/3], and ( alpha ) is a linear function of ( beta ), there exists such a ( beta ) (and consequently ( alpha )) that satisfies both ( alpha in [1/3, 2/3] ) and ( beta in [1/3, 2/3] ).Therefore, such an ( x ) exists.Wait, let me double-check. We have the equation ( k = mbeta - nalpha ). We know ( k ) is an integer in the interval ( [frac{m - 2n}{3}, frac{2m - n}{3}] ). We need to find ( alpha, beta in [1/3, 2/3] ) such that this equation holds.Since ( alpha ) and ( beta ) are both in [1/3, 2/3], ( mbeta ) is in [m/3, 2m/3] and ( nalpha ) is in [n/3, 2n/3]. Therefore, ( mbeta - nalpha ) is in [m/3 - 2n/3, 2m/3 - n/3].We already have that ( k ) is in this interval, so there exists ( alpha ) and ( beta ) such that ( k = mbeta - nalpha ). Hence, such ( alpha ) and ( beta ) exist, and consequently, such ( x ) exists.Therefore, the proof is complete.Final Answerboxed{frac{1}{3} leq {xn} leq frac{2}{3} text{ and } frac{1}{3} leq {xm} leq frac{2}{3} text{ for some real number } x.}(boxed{frac{1}{3} leq {xn} leq frac{2}{3} text{ and } frac{1}{3} leq {xm} leq frac{2}{3}})</think>