Find positive integers a,b,n that satisfy this expression

1. Nov 6, 2012

sunnybrooke

1. The problem statement, all variables and given/known data
Are there any positive integers n, a and b such that

$$96n+88=a^2+b^2$$

2. Relevant equations
3. The attempt at a solution
It resembles the Pythagorean theorem but I'm not sure how that would help me solve it. I factored the LHS

$$2^3((2^2)(3)n+11)=a^2+b^2$$

How do I proceed? Thanks.

2. Nov 6, 2012

haruspex

What can you say about perfect squares modulo 4?

3. Nov 6, 2012

sunnybrooke

I'm very bad with the whole modulus concept but I can see that a perfect square divided by 4 will always have a remainder of either 0 or 1.

4. Nov 6, 2012

haruspex

Right, so what can the sum of two squares be mod 4?

5. Nov 6, 2012

sunnybrooke

0,1 or 2, right?

6. Nov 6, 2012

haruspex

Right. Now which of those could match the LHS (mod 4)?

7. Nov 6, 2012

sunnybrooke

96 is divisible by 4 (remainder = 0). 88 is also divisible by 4 (remainder = 0). So sum of remainders (0 + 0) = 0.

8. Nov 6, 2012

SammyS

Staff Emeritus
To get the remainder of zero, means that both a and b are even. Correct ?

9. Nov 6, 2012

sunnybrooke

Yes, they'd both have to be even. Should I substitute, a = 2u, b = 2v ?

10. Nov 6, 2012

haruspex

Yes, which allows some cancellation. See if you can then repeat the logic.

11. Nov 7, 2012

sunnybrooke

96n + 88 = 4u^2 + 4v^2
24n + 22 = u^2 + v^2
(LHS= 2 mod 4 --> both u and v are odd)

24n + 22 = (2w+1)^2 + (2x+1)^2
6n + 5 = w^2 + w + x^2 + x
(LHS = 3 mod 4 --> ??)

If k^2 = 0 mod 4 then either k = 0 mod 4 or k = 2 mod 4
If k^2 = 1 mod 4 then either k = 1 mod 4 or k = 3 mod 4

Is this correct? And please correct me if my notation is wrong. Thanks.

12. Nov 7, 2012

haruspex

w2+w = w(w+1). What does that tell you about w2+w mod 2?

13. Nov 7, 2012

sunnybrooke

w(w+1)=odd*even=even --> w mod 2 = 0. same goes for x^2 + x. So the RHS is even but the LHS is odd. therefore it's not possible. Is that it?

14. Nov 7, 2012

haruspex

That's it.

15. Nov 7, 2012

sunnybrooke

Thank you so much haruspex and SammyS for breaking it down and guiding me through the problem. It helped me a lot :)