Prove that $(x+1)^p\equiv x^p+1 \mod p^2$. [closed]

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Let $p$ be a prime number and $x$ and integer such that $x^{2p}+x^p-1\equiv 0\mod p^2$.

Prove that $(x+1)^p\equiv x^p+1 \mod p^2$.

It seems to me there is no such $x$ in the first place. Any thoughts?

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2 Answers

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From $x^{2p}+x^p=1\bmod p^2$ we have

• $x(x+1)=1\bmod p$

• $x\in\Bbb{Z}/(p^2)^\times$

• $x^p+1 = x^{-p}\bmod p^2$

From $x(x+1)=1\bmod p$ we get$$x (x+1) = 1+ap\bmod p^2$$so that$$(x+1)^p=x^{-p} (1+ap)^p= x^{-p} =x^p+1\bmod p^2$$

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I think I can get part of the way. We may assume $p>5$. Suppose $x^{2p}+x^p-1\equiv 0\mod p^2$. Writing $y=x^p$, we have $$ \begin{align} y^2+y\equiv1\pmod{p^2}\\ 4y^2+4y+1\equiv5\pmod{p^2}\\ (2y+1)^2\equiv5\pmod{p^2} \end{align}$$So that $5$ is a quadratic residue modulo $p^2$. It follows from Hensel's lemma that this congruence has two solutions if $5$ is a quadratic residue modulo $p$, and no solutions otherwise. By reciprocity, $$\left(\frac5p\right)=\left(\frac p5\right)(-1)^{(5-1)(p-1)/4}=\left(\frac p5\right)$$ so that there is no solution unless $$p\equiv\pm1\pmod5.$$

Is this correct? Can it be carried further?

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