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ganeshie8
 one year ago
find all pairs of positive rational numbers \((a,b)\) such that \(\frac{a}{b}+\frac{b}{a}\) is an integer
ganeshie8
 one year ago
find all pairs of positive rational numbers \((a,b)\) such that \(\frac{a}{b}+\frac{b}{a}\) is an integer

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freckles
 one year ago
Best ResponseYou've already chosen the best response.3well an obvious pair would be (1,1)

Empty
 one year ago
Best ResponseYou've already chosen the best response.2a and b are rational numbers to begin with and not integers?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1yes, \(a\) and \(b\) are rational it seems \((1,1)\) is the only solution...

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I think set of all a, b such that a =b

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0All cases where a=b.

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[a=\frac{m}{n} \text{ and } b=\frac{r}{s} \\ \frac{a}{b}+\frac{b}{a}=\frac{(ms)^2+(rn)^2}{mrns}\] I was kinda thinking about doing something with Pythagorean triplets

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Haha right, \(\{(t,t) t\in \mathbb{Q^{+}}\}\) no other pairs ?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Hmmm it would be really nice if we weren't restricted to positive values for a and b.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1this is part of a problem where a, b happen to be positive rationals letting a,b negative simplifies the work is it ?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2I 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 \[\frac{m^2s^2 + r^2n^2}{mrns} = \frac{x^{2y}1}{x1}\] And trying to find solutions here between stuff idk

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Here is the complete problem : https://i.gyazo.com/0c248335ae5de46d24e0ac606c15be22.png

freckles
 one year ago
Best ResponseYou've already chosen the best response.3how did you reduce it down to this one

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1btw in the actual problem \(a=b\) is not a solution because the left hand side vanishes then..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1actually i wanted to show that the only integer value of `a/b + b/a` is 2 when a,b are positive are rationals

freckles
 one year ago
Best ResponseYou've already chosen the best response.3oh i see the problem came from @Astrophysics 's post it was the one posted by @mukushla

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{a}{b}+\frac{b}{a}=2 \\ a^2+b^2=2ab \\ (ab)^2=0 \\ a=b\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3but a can't be b because that one side vanishes

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so that would be mean that one thingy can't be 2

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1right, then we conclude that there are no solutions

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1it all boils down to showing that \(\frac{a}{b}+\frac{b}{a}\) cannot be an integer when \(a\ne b\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[a=\frac{m}{n} \text{ and } b=\frac{r}{s} \\ \frac{a}{b}+\frac{b}{a}=\frac{(ms)^2+(rn)^2}{mrns} = \dfrac{x^2+y^2}{xy}\] say, \(\dfrac{x^2+y^2}{xy}=t\) \( \implies \color{red}{x}^2ty\color{red}{x}+y^2=0\) \(\implies x= \dfrac{ty\pm y\sqrt{t^24}}{2}\) it follows that \(t^24\) must be a perfect square

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Wait how can \(t^24\) be a perfect square when it's \((t2)(t+2)\)?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1exactly, \(t=2\) is the only positive integer that makes \(t^24\) a perfect square (need to prove this tough)

Empty
 one year ago
Best ResponseYou've already chosen the best response.2\(n^2 \ne x*(x+4)\) 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
 one year ago
Best ResponseYou've already chosen the best response.1theres no integer way of divining that 4 to form a perfect square

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, 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 \(p^2+q^2=(p+q)^22pq=(pq)^2+2pq\) so it follows we need \((p+q)^2,(pq)^2\) divisible by \(ab\). it follows then that \((p+q)(pq)=p^2q^2\) must be divisible by \(ab\), and thus that \(p^2,q^2\) individually are divisible by \(a,b\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, where I write divisible by \(ab\) I mean by \(pq\), so we know that \(p,q\) are divisible by \(q,p\) respectively

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but if \(qp\) and \(pq\) then it follows that \(p=q\), so it must be that \(r=1/r=1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.1it follows then that \((p+q)(pq)=p^2q^2\) must be divisible by \(pq\) i didn't get that part... \((p+q)^2, (pq)^2\) are divisible by \(pq\) implies \((p+q)^2(pq)^2\) is divisible by \(pq\) right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well clearly since \((p+q)^2,(pq)^2\) are divisible by \(pq\), then \((p+q)^2(pq)^2\) is divisible by \(p^2q^2\), so \((p+q)(pq)=p^2q^2\) is divisible by \(pq\). and then since \(pq\) divides \(p^2+q^2,p^2q^2\) we can take integer combinations: $$p^2+q^2+p^2q^2=2p^2\\p^2+q^2(p^2q^2)=2q^2$$ so we have that \(pq\) divides \(2p^2,2q^2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and we can conclude that \(p2q\) and \(q2p\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1well clearly since \((p+q)^2,(pq)^2\) are divisible by \(pq\), then \((p+q)^2(pq)^2\) 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 \(pq\) have no factors in common right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no, 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
 one year ago
Best ResponseYou've already chosen the best response.1omg! yes that works beautifully!!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1i see you're working it using pure divisibility arguments i think we can finish off the proof early by considering the quadratic : \[r+\frac{1}{r}=t \implies r = \dfrac{t\pm\sqrt{t^24}}{2}\] that right hand side is irrational unless \(t=\pm 2\) \(\implies r=\pm 1\implies a=\pm b\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.2\(n^2 \ne x*(x+4)\) sorta seems intuitive for some reason to me because the values are too large and basically consecutive but idk how to reason this out.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\(n^24 = k^2 \implies (n+k)(nk)=4\) right hand side has finite number of factors...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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
 one year ago
Best ResponseYou've already chosen the best response.1Looks really neat! thnks! :)
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