I am having some trouble with the following question:
Let $m,n$ be a integers such that $m^2 + n^2 \equiv_{4}{0}$.
Prove:
If $7n + 5m =2$ then $(n,m) = 2$.
I was able to show that $m, n$ are both even.
Im not so sure how to continue. Any hints will be useful.
Alternative approach.
If $x$ is odd, then $x^2 \equiv 1\pmod{4}$. By the constraints of the problem, you can therefore conclude that
$$\text{both} ~m,n~ \text{are even}.\tag1 $$
Further, as a general rule, if you take any linear combination $Am + Bn = C$, then the gcd$(m,n) = d$ must be a divisor of $C$. This follows, since $d|m$ and $d|n$ implies that $d$ divides any linear combination involving $m,n$.
Therefore
$$\text{gcd}(m,n) ~\text{is a divisor of} ~2. \tag2 $$
The conclusion follows by putting (1) and (2) above together.
Hint : For all integer $n$, $n^2\equiv_4 0, 1$. So $n, m$ should be all even. Then?
Since $5\not\mid n$, so $(n, m)=(n, 5m)=(n, 5m+7n)=(n, 2)=2$.
I was able to show that m,n are both even.
That's a great start. This means that $2$ divides both $m$ and $n$, hence divides their gcd $d=(m,n)$.
Now, $7n+5m=2$ implies that $d$ divides $2$, so $d=2$. Indeed, $d$ divides both $n$ and $m$, so it has to divides $7n$ and $5m$, and their sum $7n+5m=2$.
