Prove that $\gcd(a,b) \leq \gcd(2a + b, a - 4b)$

Question

It is given that $a, b , 2a + b, a - 4b$ are all non zero integers.

I tried to work a contradiction by assuming that $\gcd(a,b) > \gcd(2a + b, a - 4b)$ and then using the euclidean algorithm but I couldn't see where to go from there.

Answer 1

Let $d$ be a common divisor of $a$ and $b$, i.e., $a=da_1$ and $b=db_1$. Hence, we have $2a+b = d(2a_1+b_1)$ and $a-4b = d(a_1-4b_1)$, which means $d$ is also a command divisor of $2a+b$ and $a-4b$. Hence, $\gcd(a,b) \leq \gcd(2a+b,a-4b)$.

Answer 2

It follows from the simple observation that every common divisor of $a$ and $b$ is a common divisor of $2a + b$ and $a - 4b$.

$\gcd(a,b)$ is a common divisor of $a$ and $b$.

$\gcd(2a + b, a - 4b)$ is the largest common divisor of $2a + b$ and $a - 4b$.

Therefore, $\gcd(a,b) \leq \gcd(2a + b, a - 4b)$.

(Actually, $\gcd(a,b)$ divides $\gcd(2a + b, a - 4b)$.)