Can anyone prove these two following theorems:
1)
$\gcd(a,b) = \gcd(a,b+ak) $
2)
If $\gcd(a,bc) = 1$ then $\gcd(a,b)=\gcd(a,c) = 1$
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Denote $$s:=\gcd(a,b)$$ $$t:=\gcd(a,b+ak)$$
$t$ is a common divisor of $a$ and $b+ak$, so there are integers $u,v$ with $ut=a$ and $vt=b+ak$. This implies $(v-ku)t=vt-kut=b+ak-ak=b$, hence $t|b$ implying that $t$ is a common divisor of $a$ and $b$. We can conclude $s|t$.
$s$ is a common divisor of $a$ and $b$, so there are integers $x,y$ with $xs=a$ and $ys=b$. This implies $(y+kx)s=ys+kxs=b+ak$, hence $s$ is a common divisor of $a$ and $b+ak$. We can conclude $t|s$.
Because of $s|t$ and $t|s$ we can conclude $s=\pm t$.
If we only consider non-negative numbers, the $\gcd$ will be non-negative as well. In this case, we get $s=t$, as claimed.
For exercise $2$ , we assume that $a$ and $b$ are not coprime or that $a$ and $c$ are not coprime. Then, there exists a prime $p$ with $p|a$ and $p|b$ or there exists a prime $p$ with $p|a$ and $p|c$. In both cases, the prime $p$ divides $a$ and $bc$ contradicting $(a,bc)=1$
Peter
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