Proving a | ($b^2$ + $c^2$) if a | b and a | c

Question

To prove the statement above, I made this attempt:

Let a, b, c ∈ ℤ, and a | b and a | c. Then ∃ j, k ∈ ℤ such that aj = b and ak = c. Then

$b^2$ + $c^2$ = $(aj)^2$ + $(ak)^2$ = $a^2$($j^2$ + $k^2$).

The problem is I need to show a(integer), not $a^2$. If I remove the power from a, I get a $\sqrt{(j^2 + k^2)}$, which is not necessarily an integer,

Any idea what I'm missing?

Thank you!

Answer 1

$$ a^2(j^2 + k^2) = a \left( a(j^2 + k^2) \right) \text{.} $$

( I probably would have gone with \begin{align*} \frac{b^2 + c^2}{a} &= \frac{b^2}{a} + \frac{c^2}{a} \\ &= \left( \frac{b}{a} \right)b + \left( \frac{c}{a} \right) c \\ & \in \mathbb{Z} \text{.)} \end{align*}