Direct proof of divisibility

Question

I need to directly prove, that $$7∣4a→7∣a$$

where $a\in\mathbb Z$.

Since this is a direct proof, we need to assume that $4a$ is divisible by 7, which means that $$4a=7k$$ where k is an integer. From this, we need to arrive at $a=7k$ I've tried several methods, but to me, it seems impossible to refine $4a=7k$ into $a=7k$. Which can be shown if we divide both sides by $4$. What am I missing?

Answer 1

Suppose $4a=7k$ for some $k$. Use Bézout's identity: we have $1=2\cdot 4- 7$, so, multiplying both sides by $a$, we get $$a=2\cdot 4a-7a=7(2k-a).$$

Answer 2

Since $7\mid 4a$, then we can write, that $7\mid 21a - 5 \times 4a$, because both $21$ and $4a$ are divisible by $7$, which leads to $7\mid a$.