Question

find all pairs of positive rational numbers \((a,b)\) such that \(\frac{a}{b}+\frac{b}{a}\) is an integer

Answer 1

well an obvious pair would be (1,1)

Answer 2

a and b are rational numbers to begin with and not integers?

Answer 3

yes, \(a\) and \(b\) are rational
it seems \((1,1)\) is the only solution...

Answer 4

Well what about (2,2) ?

Answer 5

I think set of all a, b such that a =b

Answer 6

All cases where a=b.

Answer 7

;)

Answer 8

\[a=\frac{m}{n} \text{ and } b=\frac{r}{s} \\ \frac{a}{b}+\frac{b}{a}=\frac{(ms)^2+(rn)^2}{mrns}\]
I was kinda thinking about doing something with Pythagorean triplets

Answer 9

Haha right, \(\{(t,t)| t\in \mathbb{Q^{+}}\}\)

no other pairs ?

Answer 10

Hmmm it would be really nice if we weren't restricted to positive values for a and b.

Answer 11

this is part of a problem where a, b happen to be positive rationals

letting a,b negative simplifies the work is it ?

Answer 12

I don't know it just gives more freedom to play around haha. I was thinking of the geometric series and trying to figure out how to make that work. Specifically taking @freckles and putting in something like

\[\frac{m^2s^2 + r^2n^2}{mrns} = \frac{x^{2y}-1}{x-1}\]

And trying to find solutions here between stuff idk

Answer 13

Here is the complete problem :
https://i.gyazo.com/0c248335ae5de46d24e0ac606c15be22.png

Answer 14

how did you reduce it down to this one

Answer 15

https://i.gyazo.com/68131a72247a651646b4589de2101e8d.png

Answer 16

btw in the actual problem \(a=b\) is not a solution because the left hand side vanishes then..

Answer 17

ah ok

Answer 18

actually i wanted to show that the only integer value of `a/b + b/a` is 2 when a,b are positive are rationals

Answer 19

oh i see the problem came from @Astrophysics 's post
it was the one posted by @mukushla

Answer 20

yes.. that one..

Answer 21

\[\frac{a}{b}+\frac{b}{a}=2 \\ a^2+b^2=2ab \\ (a-b)^2=0 \\ a=b\]

Answer 22

but a can't be b because that one side vanishes

Answer 23

so that would be mean that one thingy can't be 2

Answer 24

right, then we conclude that there are no solutions

Answer 25

it all boils down to showing that \(\frac{a}{b}+\frac{b}{a}\) cannot be an integer when \(a\ne b\)

Answer 26

\[a=\frac{m}{n} \text{ and } b=\frac{r}{s} \\ \frac{a}{b}+\frac{b}{a}=\frac{(ms)^2+(rn)^2}{mrns} = \dfrac{x^2+y^2}{xy}\]

say, \(\dfrac{x^2+y^2}{xy}=t\)

\( \implies \color{red}{x}^2-ty\color{red}{x}+y^2=0\)

\(\implies x= \dfrac{ty\pm y\sqrt{t^2-4}}{2}\)

it follows that \(t^2-4\) must be a perfect square

Answer 27

Wait how can \(t^2-4\) be a perfect square when it's \((t-2)(t+2)\)?

Answer 28

exactly, \(t=2\) is the only positive integer that makes \(t^2-4\) a perfect square (need to prove this tough)

Answer 29

\(n^2 \ne x*(x+4)\) sorta seems intuitive for some reason to me because the values are too large and basically consecutive but idk how to reason this out.