Am I proving the solutions to congruence $x^2 \equiv x \pmod{p^k}$ are $x$ such that $x \equiv 0$ or $1\pmod{p^k}$ correctly?

Question

Am I proving the following problem correctly?

Prove that the olutions to the congruence $x^2 \equiv x\pmod{p^k}$ are $x$ such that $x \equiv 0$ or $1\pmod{p^k}$, with $p, k \in \mathbb{P}$.

Dividing both sides of the congruence by $x$ gives:

$x \equiv 1\pmod{p^k}$ if $x$ is not 0.

If $x$ is $0$, then $0^2 = 0$ and $0$ mod anything is $0$, so the congruence is also satisfied, correct?

Related. Probably a duplicate, but because I answered that one, it behooves me not to hit this with a dupehammer. — Jyrki Lahtonen, Oct 29 '21 at 04:02
Anyway, your solution is not correct. You cannot divide a congruence when you don't know whether the integer you divide with has a common factor with the modulus or not. For example, if instead of a prime power your modulus is $6$ then we have $3^2\equiv3 \pmod 6$, but $3$ is not congruent to either $0$ or $1$. THIS IMPLIES THAT ANY ARGUMENT THAT DOES NOT USE THE FACT THAT THE MODULUS IS A PRIME POWER IS WRONG. — Jyrki Lahtonen, Oct 29 '21 at 04:06
You need to be careful with divisions. You have $6\equiv 2\pmod 4$, but division by two gives an incorrect claim $3\equiv1\pmod4$. All because $\gcd(2,4)>1$. When you advance in your studies you will hear about the related concepts (zero divisors, inverse) revolving around the theme of whether cancelling a factor from an equation is logically sound or not. — Jyrki Lahtonen, Oct 29 '21 at 04:07

score 0 · Accepted Answer · edited Oct 29 '21 at 04:24

This proof is not perfect at all....

Take $x=16$ which is not equal to $0$ and take $p=2$ and $k=3$ which are prime. So according to your proof $16\equiv 1 \mod 2^3$ which is certainly wrong.

Instead of taking $x=0$ and $x≠0$ as cases, you should take $x\equiv 0\pmod{p}$ amd $x\not\equiv 0\pmod{p}$ i.e $x$ is a multiple of $p$ and $x$ is not a multiple of $p$ as case 1 and case 2.

If $x\not \equiv 0 \pmod{p}$, you can prove $x\equiv 1\pmod{p}$ by dividing both sides of congruence by $x$.

If $x \equiv 0 \pmod{p}$, there is no need of proving.

Am I proving the solutions to congruence $x^2 \equiv x \pmod{p^k}$ are $x$ such that $x \equiv 0$ or $1\pmod{p^k}$ correctly?

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