Balti Tee 1998 · Ülesanne 3

Answer: $x=14, y=27$.

Rewriting the equation as $2 x^{2}-x y+5 y^{2}-10 x y=-121$ and factoring we get:

$$(2 x-y) \cdot(5 y-x)=121 .$$

Both factors must be of the same sign. If they were both negative, we would have $2 x<y<\frac{x}{5}$, a contradiction. Hence the last equation represents the number 121 as the product of two positive integers: $a=2 x-y$ and $b=5 y-x$, and $(a, b)$ must be one of the pairs $(1,121),(11,11)$ or $(121,1)$. Examining these three possibilities we find that only the first one yields integer values of $x$ and $y$, namely, $(x, y)=(14,27)$. Hence this pair is the unique solution of the original equation.