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.
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4We want to show that if $p^a$ divides $(x-1)(x+1)$, then it divides $x-1$ or $x+1$. Note that things can break down if $p=2$. For example the congruence $x^2\equiv 1\pmod{8}$ has $4$ solutions. – André Nicolas Mar 06 '14 at 02:25
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See http://math.stackexchange.com/questions/700982/mod-problem-solving/ -- duplicate? – Barry Cipra Mar 06 '14 at 03:09
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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 $x=\pm y$.
Pedro
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