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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.

Wapiti
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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 $\,i = 1 = j.$

Bill Dubuque
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  • Thanks -- but what if $\textrm{gcd}(b,d) \neq 1$? That's why I introduced the $i,j$, to get multiples of $ad,cb$... – Wapiti Jun 27 '17 at 20:04
  • By "since $,(b,d)\mid \ldots$" above, your gcd $> 1$ in that case. – Bill Dubuque Jun 27 '17 at 20:06
  • You mean the gcd of $ad+bc$ with $bd$ will be $> 1$ if $(b,d)>1$? – Wapiti Jun 27 '17 at 20:11
  • If that's what you mean I don't think it's right, as I can construct an example where $(ad+bc,db)>1$ and $(ad+2bc,db)=1$. – Wapiti Jun 27 '17 at 20:16
  • @Wapiti Impossible since $,(iad+jbc,bd) = 1,\Rightarrow,(b,d)=1,\Rightarrow,(ad+bc,bd)=1,$ as above. – Bill Dubuque Jun 27 '17 at 20:55
  • Yes you are right, I actually messed up my question because I have not been considering $bd$ in the denominator but rather $\textrm{lcm}(b,d)$. I just put $bd$ because I thought it simplified things and didn't make a difference. – Wapiti Jun 27 '17 at 21:06
  • This is the question I meant to post: https://math.stackexchange.com/questions/2338642/prove-sum-of-numerators-is-coprime-with-denominator-obtained-by-lcm – Wapiti Jun 27 '17 at 21:19
  • @Wapiti Ok, I gave a simple proof of that case (now polished a bit). – Bill Dubuque Jun 27 '17 at 23:26