10. If $p$ is an odd prime, then $\left(a+b, \frac{a^{p}+b^{p}}{a+b}\right)=$ $\qquad$ . (where $(a, b)=1$ )
10. 1 or $p$ Hint: If $\left(a+b, \frac{a^{p}+b^{p}}{a+b}\right)=t>1$, let the prime $q \mid t$, then $q \mid a+b$, so $a \equiv-b$ $(\bmod q)$.
$$
\begin{aligned}
\text { Also } \frac{a^{p}+b^{p}}{a+b} & =a^{p-1}-a^{p-2} b+\cdots+b^{p-1} \\
& \equiv p a^{p-1}(\bmod q) \\
& \equiv 0(\bmod q) .
\end{aligned}
$$

So $q \mid p \cdot a^{p-1}$.
Also $q \mid a+b,(a, b)=1$, so $q \nmid a$, so $q \mid p$. Since $p$ is a prime, so $q=p$.
So $t=p^{\alpha}$, if $\alpha \geqslant 2$, it is easy to see
$$
\frac{a^{p}+b^{p}}{a+b} \equiv p \cdot a^{p-1}\left(\bmod p^{2}\right) \equiv 0\left(\bmod p^{2}\right) .
$$

So $p|a \Rightarrow p| b$ contradiction! So $\alpha=1$.