Japan Mathematical Olympiad

Let $n$ be a positive integer such that $n+10$ and $10n$ are both perfect squares. Then, there exists a positive integer $k$ such that $10n=k^2$. Since $k$ is a multiple of $2$ and $5$, we can write $k=10\ell$ for some positive integer $\ell$, and we have $n=10{\ell }^2$. For $\ell =1$ or $\ell =2$, we have $n+10=20$ or $50$, respectively, which are not perfect squares. Therefore, we must have $\ell \ge 3$, which implies $n=10{\ell }^2\ge 10\cdot 3^2=90$.

On the other hand, since we have $90+10=100=10^2$ and $90\cdot 10=900=30^2$, $90$ satisfies the condition. Therefore, the answer is $90$.