If $(a,b)=1$ then prove $(a+b, ab)=1$. 
Let $a$ and $b$ be two integers such that $\left(a,b\right) = 1$. Prove that $\left(a+b, ab\right) = 1$.

$(a,b)=1$ means $a$ and $b$ have no prime factors in common
$ab$ is simply the product of factors of $a$ and factors of $b$.
Let's say $k\mid a+b$ where $k$ is some factor of $a$.
Then $ka=a+b$ and $ka-a=b$ and $a(k-l)=b$.
So $a(k-l)=b, \  a\mid a(k-1)$ [$a$ divides the left hand side] therefore $a\mid b$ [the right hand side].
But $(a,b)=1$ so $a$ cannot divide $b$.
We have a similar argument for $b$.
So $a+b$ is not divisible by any factors of $ab$.
Therefore, $(a+b, ab)=1$.
Would this be correct? Am I missing anything?

Intuitively, if a factor divides $a$ but not $b$ then it divides $ab$ and $a^2$ but not $b^2$, hence not $a^2+b^2$.
More formally, notice that if $a$ and $b$ are coprime then they are also coprime to $a+b$ and hence $ab$ is coprime to $(a+b)^2 = a^2 + 2ab+b^2$.
