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I can't do this last question of my homework that's due in tomorrow. Can anyone hint me on what to do?

Suppose $p$ is prime and $k$ is a positive integer

Show that if $p$ is odd and $x$ is an integer such that $x^2\equiv 1\mod p^k$ , then $x\equiv \pm 1\mod p^k$

abiessu
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John
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2 Answers2

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We have $p^k \mid x^2 - 1 = (x+1)(x-1)$. Notice that $p$ divides at most one of $x+1$ and $x-1$ as $(x+1)-(x-1)=2$ is not divisible by $p$. It follows that one of $x+1$ and $x-1$ is in fact divisible by $p^k$, hence $x \equiv \pm1 \mod p^k$.

user133281
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If p^k divides (x^2)-1 = (x-1)(x+1), then p^k either divides (x-1) or (x+1).

Let's say p^k divides (x-1). Then x ≡ 1 mod(p^k) since any prime or integer divides 0. The same argument can be used to show that x ≡ -1 mod(p^k) for the very same reason.

user80979
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