SELECTION TESTS FOR THE 2019 BMO AND IMO

Clearly, $n=1$ satisfies the required condition. We now proceed to rule out all integers $n>1$.

If $n$ is odd, $n\ge 3$, then $n^{n+1}+n-1$ falls strictly between the squares of two consecutive integers, $${\left(n^{\left(\frac{n+1}{2}\right)}\right)}^2<n^{n+1}+n-1<{\left(n^{\left(\frac{n+1}{2}\right)}+1\right)}^2,$$ so it is not the square, and hence all the less the sixth power of an integer.

Similarly, if $n\equiv 2(\mathrm{mod}3)$, then $n^{n+1}+n-1$ falls strictly between the cubes of two consecutive integers, $${\left(n^{\frac{(n+1)}{3}}\right)}^3<n^{n+1}+n-1<{\left(n^{\frac{(n+1)}{3}}+1\right)}^3,$$ so it is not the cube, and hence all the less the sixth power of an integer.

If $n\equiv 0(\mathrm{mod}3)$, then $n^{n+1}+n-1\equiv -1(\mathrm{mod}3)$, so it is not the square, and hence all the less the sixth power of an integer.

Finally, to rule out the only case left, $n\equiv 4(\mathrm{mod}6)$, notice that $$n^{n+1}+n-1\equiv (-1{)}^{n+1}-1-1\equiv -3(\mathrm{mod}n+1).$$ Since $n+1\equiv 5(\mathrm{mod}6)$, it has a prime divisor $p\equiv 2(\mathrm{mod}3)$, $p>3$, so $-3$ is not a quadratic residue modulo $p$. Consequently, $n^{n+1}+n-1$ is not a quadratic residue modulo $p$, and hence all the less the square of an integer, let alone the sixth power of one such.