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 :)
 A: As you saw, we have $2(4a+b)-(a+2b)=7a$. Also, $4(a+2b)-(4a+b)=7b$. 
So any common divisor of $4a+b$ and $a+2b$ is a common divisor of $7a$ and $7b$. 
Since $\gcd(a,b)=1$, there are integers $x$ and $y$ such that $ax+by=1$. It follows that $(7a)x+(7b)y=7$, so any common divisor of $7a$ and $7b$ divides $7$. 
Remark: We have shown that if $a$ and $b$ are relatively prime, then no positive integer other than $1$ or $7$ can be the gcd of $4a+b$ and $a+2b$. 
Each of these can arise. For gcd equal to $1$, let $a=b=1$. For gcd equal to $7$, let $a=1$ and $b=3$. 
A: $7$ 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 $7$.
A: Since $\gcd(a,b)=1$, there are $x,y\in\mathbb{Z}$ so that $ax+by=1$.
$$
\begin{align}
7
&=\overbrace{[2\color{#C00000}{(4a+b)}-\color{#00A000}{(a+2b)}]}^{\large7a}\,x+\overbrace{[4\color{#00A000}{(a+2b)}-\color{#C00000}{(4a+b)}]}^{\large7b}\,y\\[4pt]
&=(2x-y)\color{#C00000}{(4a+b)}+(4y-x)\color{#00A000}{(a+2b)}
\end{align}
$$
Therefore, $\gcd(4a+b,a+2b)\mid7$.
