<s>[INST] Prove that if the $\gcd(a,b)=1$, then $\gcd(a+b,ab)=1$ I need to prove that: 
If $\gcd(a,b)=1$, then $\gcd(a+b,ab)=1.$
So far I used what's given so I have:
$ax+by=1$ (I wrote the gcd of a and b as a linear combination)
and 
$(a+b)u+ab(v)=1$ (I also wrote this as a linear combination)
where do I go from here?
 [/INST] We're given $\gcd(a,b) = 1$, and from Bézout's identity we have $\gcd(a,b) = 1 \iff \exists x,y: ax + by = 1$ for integer $x$ and $y$. 
$1 = ax + by = ax + bx - bx + by = (a + b)x + b(y - x)$, so $\gcd(a+b,b) = 1$.  Likewise, $\gcd(a+b,a)=1$ because $(a + b)y + a(x - y) = 1$   
$$ \begin{align}
1 &= ((a + b)x + b(y - x))((a + b)y + a(x -y))
\\&= (a+b)(a+b)xy + (a+b)a(x-y) + (a+b)b(y-x)y + ab(y-x)(x-y) 
\\&= (a+b)[(a+b)xy + a(x-y) + b(y-x)] + ab[(y-x)(x-y)]
\end{align}$$
So $\gcd(a+b,ab) = 1$.
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