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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!

Bill Dubuque
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1 Answers1

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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 $Z^p=b$.

reuns
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  • Thank you! It helped. –  Dec 04 '20 at 06:25
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    How do you get from the existence of $Z_n$ to the existence of $Z_{n+1}$? The derivative of $Z_n^p-b$ with respect to $Z_n$ is $pZ_n^{p-1}$ which is $0 \mod p$. So you can't apply Hensel's lemma. – jMdA Mar 01 '21 at 02:07
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    @jMdA Multiply by $Z_1^{-p}$ to get that $Z_1^{-p}(X^p+1) \in 1+p\Bbb{Z}_p$, then show that (for odd prime) $Z_1^{-p}(X^p+1) \in 1+p^2\Bbb{Z}_p=(1+p\Bbb{Z}_p)^p$. – reuns Mar 01 '21 at 04:52
  • That is to say if $r\equiv 1\bmod p^2$ then its order $\bmod p^{n+1}$ divides $p^{n-1}$, so it is $=s^p\bmod p^{n+1}$ for some $s$. – reuns Mar 01 '21 at 05:12