Basic congruence help

Question

If I have that $a \equiv b$ mod $m$, then how do I show that $4a \equiv 4b$ mod $m$?

I understand for $4a \equiv 4b$ mod $m$ that must mean $m|(4a-4b)$, but I don't unsterstand how I would prove it.

Any help would be fantastic.

Answer 1

It's just definition. Try to follow and prove the following:

$$a=b\pmod m\implies \exists\,k\in\Bbb Z\;\;s.t.\;\;a=b+km\implies$$

$$\implies4a = 4b +(4k)m\implies 4a=4b\pmod m$$

Answer 2

Since $a \equiv b \pmod m$, we know that $m$ divides $a-b$. I.e., $a-b = mk$ for some integer $k$. Clearly, $m$ also divides $ 4a - 4b = 4(a-b) = 4(mk) = 4k(m)$. This means $$4a-4b \equiv 0 \pmod m \iff 4a \equiv 4b \pmod m$$