Find all positive integers $m$ that satisfies: $a^2 \equiv 0\mod{m} \Rightarrow a \equiv 0 \mod{m}$, $\forall a \in \mathbb{Z}$

Question

I could prove that this is statement is true if $m=1$ or if $m$ is prime.

However, I couldn't prove that the statement doesn't hold if $m>1$ is not prime.

Any hints?

@Arthur He's trying to prove that for all $m$ not prime, the statement doesn't hold (since he wants to find $m$ for which it's true) — Oussama Boussif, May 28 '17 at 23:24
@OussamaBoussif I didn't write my answer because I knew this question was a duplicate, but I couldn't find it. However, someone else found it. — Xam, May 28 '17 at 23:37
@Xam Ah okay, because I was interested in the proof as I couldn't myself prove it ^^ — Oussama Boussif, May 28 '17 at 23:41

score 0 · Answer 1 · answered May 28 '17 at 23:24

0

Prime isn't what you're looking for, it's only a subset of what you want. Consider $m = 6$. Then,

$1^2 \equiv 1$, $2^2 \equiv 4$, $3^2 \equiv 3$, $4^4 \equiv 2$, and $5^5 \equiv 1$. Thus, the conditions are satisfied, but $6$ is obviously not prime.

answered May 28 '17 at 23:24

Duncan Ramage

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Find all positive integers $m$ that satisfies: $a^2 \equiv 0\mod{m} \Rightarrow a \equiv 0 \mod{m}$, $\forall a \in \mathbb{Z}$

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