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24 divides m if n^2 -m and n^2 +m are perfect squares

  1. May 31, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that if n2 + m and n2 - m are perfect squares then m is divisible by 24.

    2. Relevant equations
    This problem comes from Larson's Problem-solving Through Problems in the section on modular arithmetic. It is the third part of a four-part problem, with the previous two parts of the problem being to prove that if x3 + y3 = z3 has a solution in integers x, y, and z then one of them is divisible by seven, and the second part to prove that if 2n + n2 is prime then n is congruent to 3 mod6. Since they're part of the same problem I assume that they are somehow relevant.

    3. The attempt at a solution

    I want to show that m is divisible by both 3 and 8. By the Chinese Remainder Theorem, I can set up the following:

    m ≡ a(mod3)
    m ≡ b(mod8)

    And I assume that what I want to do is show that the two expressions can only be perfect squares if a = b = 0.

    I don't know if any of the following actually brings me any closer to a solution, but I feel like they might be pointing in the right direction:

    I can set up the following relations:

    n2 + m ≡ n2(mod3) + a(mod3)
    n2 + m ≡ n2(mod8) + b(mod8)
    - m ≡ n2(mod3) - a(mod3)
    - m ≡ n2(mod8) - b(mod8)

    And further I can mod n out so that
    n2(mod3) + a(mod3)[a(mod3)](modn), etc

    I can also multiply n2 + m and n2 - m to get n4 - m 2, which gets

    n4 - m 2 ≡ (-a2mod3)(modn)
    n4 - m 2 ≡ (-b2mod8)(modn)

    All of which I think might be useful in establishing a = b = 0, but I can't think of how to get there.
  2. jcsd
  3. Jun 1, 2016 #2
    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.
  4. Jun 3, 2016 #3


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    What do you know about perfect squares modulo 8 and modulo 3? What are the possible differences between two perfect squares modulo 8 and 3?
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