Proof for $ax + by \equiv 0 \pmod c$ implies $x\equiv 0 \pmod c$ and $y\equiv 0 \pmod c$ when $a,b$ are co-prime

Question

I believe this is true, but I'm not sure if I'm making a mistake in my rather informal proof below.

Proof sketch: First, $ax+by \equiv 0\pmod c \Leftrightarrow \exists k \in \mathbb{Z}, ax+by = kc$.

Next, since $a,b$ are co-prime, we know that the only solutions to $ax+by = kc$ must be multiples of the Bezout coefficients $(s,t)$ such that $as + bt = 1$. This means that $x = s(kc)p$ and $y=t(kc)q$ for some $p,q\in \mathbb{Z}$, which implies $x\equiv 0 \pmod c$ and $y\equiv 0\pmod c$.^{(This is the part I'm unsure about.)}

Am I making any mistakes or do you see an easier way to prove this?

Answer 1

The claim is not true.

E.g. $ 2 \times 1 + 3 \times 1 \equiv 0 \pmod{5}$.

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Do you see where you made a reasoning error?