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I have seen people use this without proof as a well known fact.

Can someone give a proof or a reference?

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Do you know the Vieta root jumping technique? – Calvin Lin Jan 12 '13 at 0:13
found it on Wikipedia, they have this as an example. copy the comment as an answer. – Bojan Serafimov Jan 12 '13 at 0:21
up vote 2 down vote accepted

Do you know the Vieta root jumping technique?

Here is a problem for you to try, using the same technique.

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can't read the problem – Bojan Serafimov Jan 12 '13 at 0:31
@BojanSerafimov Hm, what issue are you having? Try expanding your screen, that might help. The question asks you to determine pairs of non negative integers from 1 to 100, such that $ n \mid m^2 -1, m \mid n^2 -1 $. – Calvin Lin Jan 12 '13 at 0:34
I tried firefox and chrome, both fullscreen, everything's there, just the text of the problem is missing. – Bojan Serafimov Jan 12 '13 at 0:37
@BojanSerafimov Use the idea of Vieta root jumping. Given a solution $(n, m)$, try and reduce it to something 'smaller'. – Calvin Lin Jan 12 '13 at 1:17
@BojanSerafimov It looks like it might be a Windows cache issue. Try removing toolbars or using IE 9 or 10, instead of firefox 20.0. – Calvin Lin Jan 14 '13 at 16:11

I assume that both $a,\,b\in\mathbb{N}$.

Consider for $k\in\mathbb{N}$ $a^2+b^2+1=kab$.

Note that $k\neq 1,2$ because

$a^2+b^2+1\geq 2ab+1>2ab>ab$.

I was hoping to do something similar for $k>3$ also but it doesn't work...

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I thought you just gave me a hint xD The same cant be done, because $a^2 + b^2$ can be much greater than $ab$, for example when $b=1$ – Bojan Serafimov Jan 11 '13 at 23:57

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