Proving a statement similar to the Fermat's theorem using modular arithmetic.

Question

I have to prove that if m > 1 and not a prime, then $\exists a,b,c \in \mathbb{Z}$ such that $c \not= 0 (\mod m)$, $ac = bc (\mod m)$, but $a \not = b (\mod m)$.

I am sorry I don't know how to put the three lines for the modular arithmetic symbolic representation.

Anyway, I tried assuming that $a-b$ and $c$ are two prime numbers, but then this statement would only be valid for those m that are formed but two primes.

I understand that I have to prove that $m \not| c \wedge m \not|(a-b) \wedge m | c(a-b)$

Can someone help me with this? I can't seem to find the right integers.

Edit:

Since $m = (p_1)(p_2...p_n)$ being $> 1$ and not prime by the Fundamental Thm of Arithmetic, what if I let $a = 2p_{1}$, $b = p_{1}$ and $c = (p_2....p_n)$? Then $c$ would either be $1$ or a number less than m making it not divisible by it. Same goes for $a-b$ which would be $p_1$, a prime. Is this reasoning valid?

Answer 1

$m$ is not prime so he can be factorized $m=cx$.

$c\neq 0$(mod m).

Now take $a=x$ and $b=0$ then $ac=xc=0=0\cdot c$ (mod m).