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Exercise 1.2 of Washington's book "Introduction to Cyclotomic Fields" says

Suppose $p \equiv 1$ (mod $3$). Using the fact that $\mathbb{Z}_p$ contains the cube roots of unity, show that $x^p+y^p \equiv z^p$ (mod $p^n$), $p \not| xyz$, has solutions for each $n \ge 1$.

For instance, let $p=7$. In $\mathbb{Z}_7$, cube roots of unity are $$1,~ 2 + 4\cdot 7 + 6\cdot 7^2 + 3\cdot 7^3 + \cdots,~ 4 + 2\cdot 7 + 0\cdot 7^2 +3\cdot 7^3 + \cdots.$$ Also let $n=2$. Then $x^7+y^7 \equiv z^7$ (mod $7^2$) has a desired integer solution $(x,y,z)=(1,30,31)$.

But I have no idea how to use the existence of third roots of unity to show the existence of certain integer solution.

Thank you in advance.

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1 Answer 1

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Let $\omega$ be a nontrivial cube root of 1. Then $1+\omega+\omega^2=0$, so that $$ \omega^p+(\omega^2)^p=\omega+\omega^2=-1=(-1)^p $$

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