<s>[INST] 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?
 [/INST] By contradiction, let $d>1$ divide both $ab$ and $a^2+b^2$. Then $d$ has a prime divisor $p$, which divides both $ab$ and $a^2+b^2$.
Then, $p|a(a+b) - ab = a^2$
Since $p$ is a prime which divides $a^2$ $\implies$ that $p|a$.
And $p|b(a+b) - ab = b^2$ $\implies$ $p|b$.
Hence $p|a$ and also, $p|b$, so $p$ divides $gcd(a,b) = 1$. However, this forces $p = 1$, which is a contradiction, since 1 isn't a prime.
Thus the only positive common divisor of $ab$ and $a^2+b^2$ is $1$, and hence they are coprime. 
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