Vtyjkyuk

I tried to use Euclidian Algorithm to find this answer. But I am just ended with some complicated expression with no meaning whatsoever. help!

Solution #$1$ . . .

Suppose $d$ is a common divisor of $a$ and $b$.

Then $d|a$ and $d|b$ implies $d|(a+b)$, so $d$ is a common divisor of $a$ and $a+b$.

Suppose $d$ is a common divisor of $a$ and $a+b$.

Then $d|a$ and $d|(a+b)$ implies $d|bigl((a+b)-abigr)$, so $d$ is a common divisor of $a$ and $b$.

Thus, the common divisors of $a,b$ are the same as the common divisors of $a,a+b$.

It follows that $gcd(a,a+b)=gcd(a,b)$.

In particular, if $a,b$ are coprime, so are $a,a+b$.

Solution #$2$ . . .

Let $a,b$ be coprime integers.

We want to show that $a,a+b$ are coprime.

Since $gcd(a,b)=1$, we can write $ax+by=1$, for some integers $x,y$.

Let $r=x-y$, and let $s=y$.$;$Then
$$ar+(a+b)s=a(x-y)+(a+b)y=ax+by=1$$
hence, since $ar+(a+b)s=1$, it follows that $gcd(a,a+b)=1$.

Therefore $a,a+b$ are coprime, as was to be shown.

Consider that $gcd left( a, a + b right) = d$. Then, $d | a$ and $d | a + b$. Therefore, $d | left( a + b - a right) = b$. However, $gcd left( a, b right) = 1$ (since they are coprime). Therefore, $d = 1$.

So, we can conclude that $gcd left( a, a + b right) = 1$.

It's 1. If $dk = a$ and $dl = a + b$, then $b = dk - dl = d(k-l)$. So $d$ divides $b$ as well.