Question

Let a, b, c be integers such that gcd(a, b, c)= 1. Find gcd (a+b, b+c, c).

Answer 1

Task.

Let $a,b,c$ be integers such that $gcd(a,b,c)=1$. Find $gcd(a+b,b+c,c)$.

Solution. Recall that $gcd(a,b,c)=1$ if and only if there exist integer numbers $x,y,z$ such that

$ax+by+cz=1.$

We claim that $gcd(a+b,b+c,c)=1$ as well. For this it siffices to find numbers $x^{\prime},y^{\prime},z^{\prime}$ such that

$(a+b)x^{\prime}+(b+c)y^{\prime}+cz^{\prime}=1.$

From identity

$ax+by+cz=1$

we get

$1$ $=ax+by+cz$

$=ax+bx-bx+by+cz$

$=(a+b)x+b(y-x)+cz$

$=(a+b)x+b(y-x)+c(y-x)-c(y-x)+cz$

$=(a+b)x+(b+c)(y-x)-c(y-x)+cz$

$=(a+b)\cdot\underbrace{x}_{x^{\prime}}\ +(b+c)\cdot\underbrace{(y-x)}_{y^{\prime}}\ +c\cdot\underbrace{(z-y+x)}_{z^{\prime}}.$

So we can put

$x^{\prime}=x,\qquad y^{\prime}=y-x,\qquad z^{\prime}=z-y+x.$

Answer. $gcd(a,b,c)=1$