1. Given that $n$ and $s$ are integers, if for any integer $n$, the equation $x^{2}-8 n x+7^{s}=0$ has no integer solutions, the set of all such numbers $s$ is $\qquad$ .
1. Set of even numbers.

If $\Delta=64 n^{2}-4 \cdot 7^{s}=4\left(16 n^{2}-7^{s}\right)$ is a perfect square, let $16 n^{2}-7^{s}=m^{2}$, then $m^{2} \equiv 0,1$ or $4(\bmod 8)$. And $16^{2}-7^{s}=16 n^{2}-(8-1)^{s} \equiv-(-1)^{s}(\bmod 8)$, for $s$ being odd, the original equation has integer solutions. If the original equation has no integer solutions, then $s$ must be even.