Prove the gcd $(4a + b, a + 2b) $ is equal to $1$ or $7$. So in the question it says to let $a$ and $b$ be nonzero integers such that $\gcd(a,b) = 1$.
So based on that I know that $a$ and $b$ are relatively prime and that question is basically asking if the GCD divides $7$.
So I tried to prove through the Euclidean algorithm but I think I messed up.
$a+2b = 2\times(4a+b) - 7a$
$4a+b = \frac{-4}{7}\times(-7b) +a$
$7b = 0\times a +7b$
$a = 0\times 7b +a$ 
$7b = 0\times a + 7b$
and so on and so on. 
The fact that it is repeating tells me that I am doing something wrong.
Thanks you in advance for all of your help :)

$7$ has a geometric interpretation here.
Take integers $u$ and $v$ such that $av-bu=1$. Then the parallelogram in $\mathbb Z^2$ defined by the vectors $(a,b)$ and $(u,v)$ has area $1$.
The linear transformation $T(x,y)= (4x + y, x + 2y)$ sends that parallelogram to another parallelogram having area $7=\det T$.
Hence $7=(4a + b)(u+2v)-(a + 2b)(4u+v)$. This proves that $\gcd(4a + b, a + 2b)$ is a divisor of $7$.
