SELECTION TESTS FOR THE 2019 BMO AND IMO

There is only one such integer, namely, $n=3$; it is readily checked that $3^3+1=28$ is perfect.

If $n$ is odd, then $n^n+1$ is even, so it is of the form $2^{p-1}(2^p-1)$, where $p$ and $2^p-1$ are both prime (Euler's theorem on the structure of perfect even integers). Rule out the trivial case $n=1$, to assume $n>1$, and write $n^n+1=(n+1)(n^{n-1}-n^{n-2}+\cdots -n+1)$. Since $n$ is odd, the first factor is even and the second is odd; and since $n>1$, the latter is greater than $1$ (simply rewrite it in the form $1+n(n-1)(1+n^2+\cdots +n^{n-3})$). It follows that $n+1=2^{p-1}$, so $2^p-1=2n+1$, and $n^n+1=(n+1)(2n+1)=2n^2+3n+1$ which forces $n=3$.

We now rule out the other parity of $n$; recall that the existence of perfect odd numbers is still an open question.