If $a$ and $b$ are coprime then $7a+3b$ and $2a-b$ are also coprime

Question

I am trying to prove following: If $a$ and $b$ are coprime, then $7a+3b$ and $2a-b$ are also coprime. Thanks.

Answer 1

Let $d=\gcd (7a+3b,2a-b)$ .

This means that $d \mid 7a+3b$ and $d \mid 2a-b$ so :

$$d \mid (7a+3b)+3(2a-b)=13a$$

$$d \mid 2(7a+3b)-7(2a-b)=13b$$

But $\gcd (a,b)=1$ so we must have :

$$d \mid 13$$

If $d=1$ then we're done but there are cases when $d=13$ :

For example $a=14$ , $b=15$ .

We have $ \gcd (a,b)=1$ but $2a-b=13$ and $7a+3b=143=13 \cdot 11$ so :

$$\gcd (7a+3b,2a-b)=13$$

This means that the problem is wrong .

But the reverse problem works :

If $\gcd (7a+3b,2a-b)=1$ then $\gcd(a,b)=1$ .

Let $\gcd(a,b)=d$ then because $d \mid a$ and $d \mid b$ :

$$d \mid 7a+3b$$ and also :

$$ d \mid 2a-b$$

But $\gcd (7a+3b,2a-b)=1$ so $d=1$ .

This means that $\gcd(a,b)=1$