Say I have two rational numbers $a/b$ and $c/d$ where $a,b,c,d$ are integers and $a<b$ and $c<d$, and $a$ coprime with $b$, and $c$ coprime with $d$. Assume $b,d$ are free and not necessarily coprime, and also that $d>b$ with $d$ not a power of $b$. So to add the fractions we can do $ad/bd + cb/bd$. Then $ad, cd$ are clearly not coprime with $bd$. I want to show that there exists integers $i,j$ such that $iad + jcb$ is coprime with $bd$.
I tested it on a computer and it seems true for $a,b,c,d$ less than 100. I think I can prove it, but before I spend more time on it I thought to ask if the answer already exists, or this is an instance of a well-known number theoretic result.
Hint $ $ Since $\,bd,\, iad+jcb\,$ are coprime their common divisor $\,\color{#c00}{(b,d) = 1},\,$ so, by Euclid
$\qquad (ad+bc,b) = (ad,b) = \color{#c00}{(d,b)} = 1\ $ by $\ (a,b) = 1$
$\qquad (ad+bc,d) = (bc,d) = \color{#c00}{(b,d)} = 1\ $ by $\ (c,d) = 1$
Since $\,ad+bc\,$ is coprime to $b,d$ it's coprime to $\,bd\,$ by Euler. So we can choose $\,i = 1 = j.$