Prove $\gcd(b^2+ab,a^2+2ab)=1 $ if $\gcd(a,b)=1$

Question

I want to show that for $\gcd(a,b)=1$, $a,b \in \mathbf{Z}$, $\gcd(b^2+ab,a^2+2ab)=1$.

I think the first step is something like this $$\gcd(b^2+ab,a^2+2ab)=\gcd(b(a+b),a(a+b)+ab). $$

I have two properties that can be used. There are

1/ $\gcd(a+b,c)=\gcd(a,c)$ if $b \vdots a$.

2/ $\gcd(ab,c)=\gcd(a,c)$ if $\gcd(b,c)=1$.

Please help me. Thank you.

Answer 1

Due to coprimality of $a,b$ we have $b$ coprime with $a^2+2ab$, $a$ coprime with $b^2+ab$, hence by property 2 we can simplify $$ \gcd(b^2+ab,a^2+2ab)=\gcd(b+a,a+2b). $$ Another useful property is $\gcd(m,n)=\gcd(m+n,n)$, so we continue $$ \gcd(b+a,a+2b)=\gcd(b+a,b)=\gcd(a,b)=1. $$

Answer 2

Hint

Let $d\neq 1$ a prime number s.t. $$d\mid b^2+ab=b(a+b)\quad \text{and}\quad d\mid a^2+2ab= a(a+2b).$$

• Suppose $d\mid b$. Then $\gcd(d,a)=1$, and thus $d\mid a+2b$ and thus $$a\equiv -2b\equiv 0\pmod d,$$ which contradict $\gcd(a,b)=1$. Therefore $d\nmid b$, and thus $d\mid a+b$

• I let you consider the other cases, i.e. $d\mid a+b$, $d\mid a$ and $d\mid a+2b$.