Direct proof square

Question

$4\mid 4$$k^2$

Concerns:

Is this question answered to the best of its ability or could it use more explanation. Am I missing any steps

Answer 1

The idea is there, but if you want to be more formally correct, notationwise and structurewise:

Suppose $c$ is even.

Then $\exists k \in \mathbb{Z}$ such that $c=2k$. (definition of even, or $2|c$).

Then for that same $k$: $c^2 = 2k \cdot 2k = 4k^2$, so $\exists l \in \mathbb{Z}$ such that $c^2 = 4l$ (namely $l=k^2$) so that by definition $4|c^2$.

Done: we started with the assumption $c$ is even and ended with the desired conclusion $4|c^2$ using just definitions and standard arithmetic facts.