Exercise 2.1.46* ($a^p \equiv b^p \pmod p \Rightarrow a^p \equiv b^p \pmod {p^2}$)

Question

For any prime $p$, if $a^p \equiv b^p \mod p$, prove that $a^p \equiv b^p \pmod {p^2}$.

Richard Ganaye · Accepted Answer

\begin{proof} Assume that $a^p \equiv b^p \pmod p$. By Fermat's theorem, $a^p \equiv a \pmod p$ and $b^p \equiv b \pmod p$, thus
$$a\equiv b \pmod p.$$
So  $b = a + kp$ for some integer $k$. Then, by the binomial formula,
\begin{align*}
b^p &= (a + kp)^p\
&= a^p + \binom{p}{1} a^{p-1} kp + \sum_{j=2}^p \binom{p}{j} a^{p-j} k^j p^j\
&= a^p + a^{p-1} k p^2 + K p^2,
\end{align*}
where $K = \sum_{j=2}^p \binom{p}{j} a^{p-j} k^j p^{j-2}$ is an integer. Therefore
$$b^p \equiv a^p \pmod {p^2}.$$
\end{proof}

Exercise 2.1.46* ($a^p \equiv b^p \pmod p \Rightarrow a^p \equiv b^p \pmod {p^2}$)

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Exercise 2.1.46* ($a^p \equiv b^p \pmod p \Rightarrow a^p \equiv b^p \pmod {p^2}$)

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