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Larson 3.2.16d

  1. Jan 4, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that if n^2+m and n^2-m are perfect squares, then m is divisible by 24.


    2. Relevant equations



    3. The attempt at a solution
    I found all of the squares mod 24. They are:{0,1,4,9,12,16}. We want to show that if we take any one of these as n^2, then n^2+m and n^2-m cannot be in that set. However, if we take n^2=0 and m=12, then n^2+12=n^2-12=12. What is wrong here?
    ,
     
  2. jcsd
  3. Jan 4, 2008 #2

    morphism

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    Just because some square is congruent to s modulo n it doesn't follow that everything congruent to s modulo n is a square.

    Personally I wouldn't work directly with squares mod 24, but instead with squares mod 3 and mod 4 - they're particularly straightforward. After you show that m=0 mod 3 and mod 4, proceed to show that m is necessarily even.
     
  4. Jan 4, 2008 #3
    It is even because it is congruent to 0 mod 4. What we need to show is that it is congruent to 0 mod 8.

    When I try to do that, I run into the same problem as with mod 24. n^2 could be congruent to 0 mod 8, m could be congruent to 4 mod 8, then it is possible that n^2-m and n^2+m are squares since n^2-4 = n^2+4=4=2^2 mod 8
     
    Last edited: Jan 4, 2008
  5. Jan 4, 2008 #4

    morphism

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    Wow... How silly of me! :redface:

    OK, how about this: Consider squares mod 16 instead. They are {0,1,4,9}. 2n^2 is the sum of two squares, and it's twice a square, so we get that n^2-m=n^2+m (mod 16), so that either m=0 or m=8 (mod 16), and consequently m=0 (mod 8).

    Try to fill in the details.

    (Hopefully I haven't said anything stupid this time.)
     
  6. Jan 4, 2008 #5
    Yes, that works. The only possibilities for n^2 are 0,1,9.

    2*0=0+0 and 2*1 = 1+1 and 2*9=2=1+1

    so m = 0,0,8 respectively

    I am not sure that's really an elegant proof because I just checked all the cases mentally, though...
     
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