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The puzzle is as follows:

Problem: Find all non-negative integer solutions to $4ab-a-b=c^2$

My Progress:

There is, of course, the trivial solution of $a=b=c=0$, and I suspect there are no more (though I would not be surprised if I was wrong).

Knowing this, an equivalent problem is to find the solutions to $(4a-1)|(c^2+a)$ or, in modular arithmetic, $c^2\equiv-a~(\mathrm{mod}~(4a-1))$. In particular, the second form allows me to rule out $a=1,2,3,4,5$ as potential candidates by hand.

Now I'm stuck. Any help would be appreciated.

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  • $\begingroup$ See this. Because I answered there, I am reluctant to cast a first (and in my case also binding) vote to close this as a duplicate. $\endgroup$ – Jyrki Lahtonen Jul 23 '16 at 9:04
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Given equation can be written as $a( 4b - 1) -1/4(4b-1) = c^2 + 1/4 $

This is equivalent to $ (4a -1)(4b -1) = (4c^2 + 1) $.Use modulo arithmetic now

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  • $\begingroup$ Hm? I think I tried something like this, but stopped when I realized that there were solutions to this in every non-variable modulus I could think of. Is there some choice of modulus that I am not aware of that would solve this? $\endgroup$ – Nathaniel B Jul 23 '16 at 8:48
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    $\begingroup$ @NathanielB If $p$ is a prime factor of $4c^2+1$, then $-1$ is a QR modulo $p$, and therefore $p\equiv1\pmod 4$. What does that say about the prime factors of $(4a-1)(4b-1)$? $\endgroup$ – Jyrki Lahtonen Jul 23 '16 at 8:55

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