Talk:1988 IMO Problems/Problem 6

I just wonder if it's possible to solve this problem with Chinese Remainder Theorem

First: assuming that $GCD(a,b)=1$ .

Then quotient is always square $mod a$ and $mod b$ and is less or equal than $ab$ and is not divisible by neither $a$ nor $b$ which implies it's square of integer.

In case of $GCD(a,b) = d>1$ we can transform quotient to $d^2((a_1)^2 + (b_1)^2)/(d^2a_1b_1 + 1)$ where $a_1 = a/d$ and $b_1 = b/d$ and follow the same reasoning as above.

It's just an idea without final and rigorous proof yet.

Am I mistaken?

Help :)

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Talk:1988 IMO Problems/Problem 6