# User talk:Suvamghosh

Hi, Suvamghosh,

Thanks for your proposed solution to 2007 IMO Problems/Problem 5. Unfortunately, there's a flaw with it, so I've commented out your solution for now, but it's easy to replace if you can fix it.

Note that $(4ab-1)$ may not be prime (in fact, if the problem statement holds, then it's almost certainly not prime, as it factors as a difference of squares). So in general, $xy \equiv 0 \pmod{4ab-1}$ does not imply that $x \equiv 0 \pmod{4ab-1}$ or $y \equiv 0\pmod{4ab-1}$. This is bad, because you appear to use this implication in your solution when you move from $(4a)^2(a-b)^2 \equiv 0$ to $a-b \equiv 0$. It is okay to say that $(a-b)^2 \equiv 0$, since $4a$ is pretty clearly relatively prime to $4ab-1$, but we need some more argument here. In particular, we could (hypothetically) have $4ab-1 = n^2m$ and $a-b = nm$ or something like that. It could be the beginning of a nice solution, though.

Thanks,
Boy Soprano II 03:02, 23 November 2009 (UTC)