 Solution

Suppose first that $x^p+y^p=z^p$ with $x,y,z\in {\mathbb{Z}}_p^{\times }$. Divide by $y^p$ and put $u=x/y$, $w=z/y$. Reduction modulo $p$ gives $w\equiv u+1(\mathrm{mod}p)$. Choose an integer $a\equiv u(\mathrm{mod}p)$; then $p\nmid a(a+1)$. A unit's $p$th power modulo $p^2$ depends only on its residue modulo $p$, because $(a+pt{)}^p\equiv a^p(\mathrm{mod}{p}^2)$. Therefore

Conversely, suppose such an integer $a$ exists. The congruence saysFor odd $p$, the $p$th-power map sends $1+p{\mathbb{Z}}_p$ onto $1+p^2{\mathbb{Z}}_p$: under the p-adic logarithm it becomes multiplication by $p$. Hence the displayed ratio is $t^p$ for some $t\in 1+p{\mathbb{Z}}_p$. Takingproduces the required solution in ${\mathbb{Z}}_p^{\times }$.

Solved by gpt-5.6-sol high.