I read this thread and the author starts by stating a problem:
Suppose $p>2$ is a prime. Show that $x^p+y^p=z^p$ has a solution in $\mathbb{Z}_p^{\times}$ if and only if there exists an integer $a$ such that $p\not\mid a(a+1)$ and $$(a+1)^p=a^p+1\pmod{p^2}.$$
They claim that this is a good exercise in Hensel's Lemma and they will not prove it there. Coincidentally, I first came in touch with Hensel's lemma today, and do not know how to prove this result. Can someone help me please?
Thanks in advance!
Best Answer
If there is a solution in $\Bbb{Z}_p^\times$ then there is a solution in $\Bbb{Z}/(p^2)^\times$ which is $z^p=x^p +y^p \bmod p^2$ which gives a solution $Z^p=X^p+1 \bmod p^2$.
Since $$(Z+p^n c)^p= Z^p \bmod p^{n+1}$$ then $Z\bmod p$ determinates $Z^p \bmod p^2$, thus $Z^p = X^p+1\bmod p^2$ implies $(X+1)^p = X^p+1\bmod p^2$.
Conversely given $b=X^p+1$ and $n\ge 1$, if there is $Z_n$ such that $Z_n^p=b\bmod p^{n+1}$ then there is $Z_{n+1}\equiv Z_n \bmod p^n$ such that $Z_{n+1}^p = b\bmod p^{n+2}$, thus if $Z_1$ exists then $Z=\lim_{n\to \infty} Z_n$ is a solution to $Z^p=b$.