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

- ganeshie8

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- freckles

well an obvious pair would be (1,1)

- Empty

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

- ganeshie8

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

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## More answers

- Empty

Well what about (2,2) ?

- Loser66

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

- anonymous

All cases where a=b.

- Empty

;)

- freckles

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

- ganeshie8

Haha right, \(\{(t,t)| t\in \mathbb{Q^{+}}\}\)
no other pairs ?

- Empty

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

- ganeshie8

this is part of a problem where a, b happen to be positive rationals
letting a,b negative simplifies the work is it ?

- Empty

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

- ganeshie8

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

- freckles

how did you reduce it down to this one

- ganeshie8

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

- ganeshie8

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

- freckles

ah ok

- ganeshie8

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

- freckles

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

- ganeshie8

yes.. that one..

- freckles

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

- freckles

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

- freckles

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

- ganeshie8

right, then we conclude that there are no solutions

- ganeshie8

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

- ganeshie8

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

- Empty

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

- ganeshie8

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

- Empty

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

- dan815

|dw:1439177787318:dw|

- dan815

theres no integer way of divining that 4 to form a perfect square

- anonymous

well, if \(a,b\) are rational then \(a/b,b/a\) are rational so let's focus on rationals of the form \(r,1/r\): $$r+\frac1r=\frac{r^2+1}r$$ so lets write \(r=p/q\) in reduced form giving $$\frac{p^2/q^2+1}{p/q}=\frac{p^2+q^2}{pq}$$ so we want integers \(p,q\) such that \(pq\) divides \(p^2+q^2\). consider \(p^2+q^2=(p+q)^2-2pq=(p-q)^2+2pq\) so it follows we need \((p+q)^2,(p-q)^2\) divisible by \(ab\). it follows then that \((p+q)(p-q)=p^2-q^2\) must be divisible by \(ab\), and thus that \(p^2,q^2\) individually are divisible by \(a,b\)

- anonymous

oops, where I write divisible by \(ab\) I mean by \(pq\), so we know that \(p,q\) are divisible by \(q,p\) respectively

- anonymous

but if \(q|p\) and \(p|q\) then it follows that \(p=q\), so it must be that \(r=1/r=1\)

- anonymous

so the rational \(a,b\) must be such that \(a=b\) so that \(r=a/b=1=b/a=1/r\), meaning the solutions are precisely: $$\{(x,x):x\in\mathbb Q^+\}$$

- ganeshie8

it follows then that \((p+q)(p-q)=p^2-q^2\) must be divisible by \(pq\)
i didn't get that part...
\((p+q)^2, (p-q)^2\) are divisible by \(pq\) implies \((p+q)^2(p-q)^2\) is divisible by \(pq\) right ?

- anonymous

well clearly since \((p+q)^2,(p-q)^2\) are divisible by \(pq\), then \((p+q)^2(p-q)^2\) is divisible by \(p^2q^2\), so \((p+q)(p-q)=p^2-q^2\) is divisible by \(pq\). and then since \(pq\) divides \(p^2+q^2,p^2-q^2\) we can take integer combinations: $$p^2+q^2+p^2-q^2=2p^2\\p^2+q^2-(p^2-q^2)=2q^2$$ so we have that \(pq\) divides \(2p^2,2q^2\)

- anonymous

and we can conclude that \(p|2q\) and \(q|2p\)

- ganeshie8

well clearly since \((p+q)^2,(p-q)^2\) are divisible by \(pq\), then \((p+q)^2(p-q)^2\) is divisible by \(p^\color{red}{2}q^\color{red}{2}\)
sorry to persist, but that square is troubling me.. that is true only if \(p+q\) and \(p-q\) have no factors in common right ?

- anonymous

no, because if \(pq\) divides each individually then clearly each contributes at least one factor of \(pq\) so their product must have \((pq)^2=p^2q^2\)

- ganeshie8

omg! yes that works beautifully!!

- ganeshie8

i see you're working it using pure divisibility arguments
i think we can finish off the proof early by considering the quadratic :
\[r+\frac{1}{r}=t \implies r = \dfrac{t\pm\sqrt{t^2-4}}{2}\]
that right hand side is irrational unless \(t=\pm 2\)
\(\implies r=\pm 1\implies a=\pm b\)

- Empty

- ganeshie8

\(n^2-4 = k^2 \implies (n+k)(n-k)=4\)
right hand side has finite number of factors...

- anonymous

Well, it's essentially the same thing that @oldrin.bataku did, but it's a different way of saying it.
Let \(\frac{a}{b} = \frac{p}{q} \) in its reduced form , that is \((p,q) = 1\)
Suppose now \(\frac{p}{q} + \frac{q}{p} = n\) where \(n\) is an integer.
So, \(p^2 + q^2= npq\). Now, \(q\) divides \(npq\) and \(q^2\) . So \(q\) must divide \(p^2\), but \((p,q) = 1\) , so, \(q\) must be \(1\). Similarly, we can argue for \(p\) to show \(p\) must be \(1\).
so, \(a/b = p/q = 1\)
so whenever \(a/b+b/a\) is an integer, \(a\) and \(b\) must be equal.

- ganeshie8

Looks really neat! thnks! :)

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