# 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 :)