$q$ is prime and $k=2,3,4,\dots$ . Show that if $x\in\Bbb Z$ such that $x^2\equiv x\pmod {q^k}$ then $x\equiv 0\pmod {q^k}$ or $x\equiv 1\pmod {q^k}$

Question

Considering the or condition in this statement, would I be proving both could be true or two separate cases? Also, is there some sort of reduction required? I know there exists a formula for $x^2\equiv x\pmod {q^k}$ but something about the exponent is throwing me off, please help!

score 1 · Answer 1 · answered Feb 07 '14 at 03:34

Hint. We have $$\eqalign{x^2\equiv x\pmod{q^k}\quad &\Leftrightarrow\quad x^2-x\equiv0\pmod{q^k}\cr &\Leftrightarrow\quad x(x-1)\equiv0\pmod{q^k}\cr &\Leftrightarrow\quad q^k\mid x(x-1)\ .\cr}$$ Now can you explain why one of the statements $q^k\mid x$ and $q^k\mid x-1$ must be true?

$q$ is prime and $k=2,3,4,\dots$ . Show that if $x\in\Bbb Z$ such that $x^2\equiv x\pmod {q^k}$ then $x\equiv 0\pmod {q^k}$ or $x\equiv 1\pmod {q^k}$

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