Prove that for $\forall m, n \in \mathbb Z$ such that $[7(m + n)^2 + 2mn] \ \vdots \ 225$ then $mn \ \vdots \ 225$. 
Prove that for $\forall m, n \in \mathbb Z$ such that $[7(m + n)^2 + 2mn] \ \vdots \ 225$ then $mn \ \vdots \ 225$.

This is taken directly from an exam my friend took on $03/06/2019$. I want to ask if there are any other solutions that are more practical.
 A: We have
$$7(m + n)^2 + 2mn = 7 m^2 + 16 m n + 7 n^2 \equiv (m-n)^2 \bmod 3$$
$$7(m + n)^2 + 2mn = 7 m^2 + 16 m n + 7 n^2 \equiv 2(m-n)^2 \bmod 5$$
Therefore, $m\equiv n \bmod 15$. Write $m=15t+n$. Then
$$
7(m + n)^2 + 2mn = 30 n^2 + 450 n t + 1575 t^2 \equiv 30n^2 \bmod 225
$$
Thus,
$
30n^2 \equiv 0 \bmod 225
$
and so
$
2n^2 \equiv 0 \bmod 15
$,
which gives
$
n \equiv 0 \bmod 15
$.
Therefore, $m \equiv n \equiv 0 \bmod 15$ and $mn \equiv 0 \bmod 15^2$.
A: We have that $[7(m + n)^2 + 2mn] \ \vdots \ 225 \implies [4(7m^2 + 7n^2 + 16mn) - 225mn] \ \vdots \ 225$
$\implies (28m^2 + 28n^2 - 161mn) \ \vdots \ 225 \implies (4m^2 + 4n^2 - 23mn) \ \vdots \ 225$
$\implies [(7m^2 + 7n^2 + 16mn) + 2(4m^2 + 4n^2 - 23mn)] \ \vdots \ 225$
$\implies (15m^2 + 15n^2 - 30mn) \ \vdots \ 225 \implies (m - n)^2 \ \vdots \ 15 \implies (m^2 + n^2 - 2mn) \ \vdots \ 225$
$\implies [4(m^2 + n^2 - 2mn) - (4m^2 + 4n^2 - 23mn)] \ \vdots \ 225$
$\implies 17mn \ \vdots \ 225 \implies mn \ \vdots \ 225$
Note: Because $(7, 225) = (17, 225) = 1$, we can make simplifications like above. In addition, if $x \in \mathbb Z$ and $x^2 \ \vdots \ 15$ then we have that $x^2 \ \vdots \ 225$.
A: $$ 7 f = (7m+8n)^2 - 15 n^2 =  (8m+7n)^2 - 15 m^2  $$
For this to be divisible by $9$ and $25,$ it is first necessary that both $7m+8n$ and $8m+7n$ be divisible by both $3$ and $5,$ so that their squares are divisible by $9$ and $25.$ At this stage, we need both $15 n^2$ and $15 m^2$ to be divisible by $9$ and $25.$  It  follows that both $m,n$ are divisible by $3$ and $5.$ Thus $mn$ is divisible by $9$ and $25$
