Congruence problem

Question

Show that if $p$ is an odd prime, then the congruence $x^2\equiv1\bmod{p^a}$ has only two solutions $x\equiv1(\bmod{p^a})$ and $x\equiv-1(\bmod{p^a})$.

I'm not so sure about how to start approaching this, any suggestions please? Thank you very much.

Answer 1

It is essential that $p$ is odd. Suppose $x,y$ are both solutions, so in particlar $y\neq 0\mod p$. Note that $x^2=y^2\mod p^a$ means $p^a\mid (x+y)(x-y)$. In particular, $p\mid (x+y)(x-y)$, so $p\mid x-y$ or $p\mid x+y$. But we have that $x+y-(x-y)=2y$. Since $p\nmid y$ and $p\nmid 2$, we conclude $p$ divides exactly one of $x-y$ or $x+y$, whence $x=\pm y$.