THE Tenth STARS OF MATHEMATICS COMPETITION

The required integers are $k=3$ and $n=2$. It is readily checked that these integers satisfy the condition in the statement.

To show that there are no other such, write $2^k+10n^2+n^4=N^2$, where $N$ is a positive integer, so $(N-n^2-5)(N+n^2+5)=2^k-25$.

The latter shows that $N\ge n^2+6$ if $k\ge 5$, and of the first four positive integers, only $k=3$ yields a positive integer $n=2$, in which case $N=8$.

Henceforth, let $k\ge 5$, and notice that $n$ must be even, for otherwise $N^2=2^k+10n^2+n^4\equiv 3(\mathrm{mod}4)$, which is impossible. Let $2^m,m\ge 1$, be the highest power of $2$ dividing $n$.

If $k>2m+1$, then $2^{2m+1}$ is the highest power of $2$ dividing $2^k+10n^2+n^4=N^2$, which is again impossible.

Consequently, $k\le 2m+1$, so $2^k\le 2^{2m+1}\le 2n^2$, and $N^2=2^k+10n^2+n^4\le 12n^2+n^4<(n^2+6{)}^2$; that is, $N<n^2+6$, in contradiction with $N\ge n^2+6$ established above. This ends the proof.