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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.

FactorY
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  • Suppose that $\gcd(b^2ab,a^2+2ab)>1$, then for some prime $p$ we have $p\mid b^2+ab=b(a+b)$ and $p\mid a^2+2ab=a(a+2b)$, so we have 4 cases to consider. – richrow Sep 13 '20 at 08:27
  • Apply the first dupe's Lemma (= $3$ applications of "2/") using $,A,B,K,M = b,a+b,a,(a+2b),,$ using "1/" to deduce $, 1 = (A,K) = (A,M) =\ldots$ – Bill Dubuque Sep 14 '20 at 00:22

2 Answers2

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 $d\mid a+2b$.

Surb
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  • If d is not divisible for both b and (a+b) as well as both a and (a+2b), what will happen? I think we should consider those cases. – FactorY Sep 13 '20 at 10:33
  • @FactorY This will not happen, since $d$ is prime. – Trebor Sep 13 '20 at 10:41
  • @FactorY The idea is this: in the field $,\Bbb F_p,$ = integers $!\bmod p,$ (or in any integral domain) we have: $, x(y+x) = 0 = y(y+2x) \iff x=0=y,,$ i.e. the union of the $y$ axis and line $y=-x$ meets the union of the $x$ axis and line $y=-2x$ only at the origin $,(x,y)=(0,0).,$ The proof is easy: $,x(y+x),\Rightarrow x=0,$ or $,y = -x.,$ If $,x=0,$ then $,0 = y(y+2x) = y^2\Rightarrow, y=0.,$ If $,y = -x,$ then $, 0 = y(y+2x) = -x(x),\Rightarrow, x=0,\Rightarrow, y=-x =-0 = 0\ \ \ $ – Bill Dubuque Sep 14 '20 at 23:33
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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 $$ \gcd(b+a,a+2b)=\gcd(b+a,b)=\gcd(a,b)=1. $$

Sil
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