Larson 3.2.16d

1. Jan 4, 2008

ehrenfest

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?
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2. Jan 4, 2008

morphism

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.

3. Jan 4, 2008

ehrenfest

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
4. Jan 4, 2008

morphism

Wow... How silly of me!

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.)

5. Jan 4, 2008

ehrenfest

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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