If $\gcd(a,b)=1=\gcd(c,d)$, $b>0$, $d>0$ and $\frac {a}{b}+ \frac {c}{d}$ is an integer, show that $b=d$

Question

If $\gcd(a,b)=1=\gcd(c,d)$, $b>0$, $d>0$ and $\frac {a}{b}+ \frac {c}{d}$ is an integer, show that $b=d$.

Can someone give me hints to solve the problem? I have a feeling that we need to show that $b|d$ and $d|b$, but cant figure out how to do it. Can anyone help me? Just need some hint not the entire solution!

What do you mean by $(a,b) = 1$? Can you explain the notation? — gemspark, Aug 24 '20 at 04:29
It would be better if you add this info in the question text, @Kritesh Dhakal. — gemspark, Aug 24 '20 at 04:32
It's obvious using uniqueness of reduced fractions - see my answer in the linked dupe. — Bill Dubuque, Aug 24 '20 at 05:52

score 4 · Answer 1 · answered Aug 24 '20 at 04:35

4

Note that $$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\in\mathbb Z\implies b|(ad+bc)$$but $b|bc\implies b|ad$ and as $b$ and $a$ are co-prime, $b|ad\implies b|d$. Similarly, we get $d|b$ and thus, $b=d$.

answered Aug 24 '20 at 04:35

Anand

3,588

1

Thanks it was simple and elegant as well! – Kritesh Dhakal Aug 24 '20 at 04:43
@KriteshDhakal You're welcome :) – Anand Aug 24 '20 at 05:01

score 2 · Answer 2 · answered Aug 24 '20 at 04:39

2

You have that $ad+bc$ is divisible by $bd$. Then $ad$ must be divisible by $b$. Since $(a,b)=1$ we have that $b$ divides $d$. Similarly $bc$ is divisible by $d$, so $d$ divides $b$. Hence $b=d$.

answered Aug 24 '20 at 04:39

markvs

19,653

Thanks it was simple and elegant! – Kritesh Dhakal Aug 24 '20 at 04:43

If $\gcd(a,b)=1=\gcd(c,d)$, $b>0$, $d>0$ and $\frac {a}{b}+ \frac {c}{d}$ is an integer, show that $b=d$

2 Answers2