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24 divides m if n^2 -m and n^2 +m are perfect squares
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[QUOTE=".Scott, post: 5488662, member: 489053"] Well, you have ##a^{2}+b^{2}=2n^{2}##. Note that if a and b share any common factors, so will n. So for example, if a and b are even, n will also be even and if a and b are multiples of 3, n will also be a multiple of 3. So we can start by dividing a, b, and n by the greatest common divisor (GCD) of a and b. Working in modulo 3: n = 0, 1, 2; so ##n^{2}## = 0, 1, 1 So the square is never 2. ##a^{2}+b^{2}=2n^{2}## becomes (0+1,1+0, or 1+1) = 0 or 2 Only the 1+1=2 solution is possible. This means that none of a, b, and n are divisible by 3. So ##a^{2}-b^{2}=2m## becomes (1-1)=2m, so m=0 modulo 3. Working in modulo 2: ##a^{2}=a; b^{2}=b, 2n^{2}=0## So ##a^{2}+b^{2}=2n^{2}## becomes a+b=0 or a=b If a and b were both even, we would reduce the problem by an even GCD. So they must be both odd. Let ##a=2A+1; b=2B+1## So ##2m = a^{2}-b^{2} = (2A+1)^{2}-(2B+1)^{2} = 4A^{2}+4A+1-4B^{2}-4B-1 = 4(A^{2}+A-B^{2}-B)## So ##m = 2(A^{2}+A+B^{2}+B) = 2(A(A+1)-B(B+1))## Since A and A+1 are consecutive number, one will be even and the product will be even. Same with B. So (A(A+1)-B(B+1)) is even-even which is even. And m = 2(even) which is 0 modulo 4. So I've got m modulo 12 = 0 Not quite what you need, but I hope it helps. [/QUOTE]
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24 divides m if n^2 -m and n^2 +m are perfect squares
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