Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Can someone give a proof or a reference?

share|improve this question
1  
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

2 Answers 2

up vote 1 down vote accepted

Do you know the Vieta root jumping technique?


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

share|improve this answer
    
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
1  
@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
1  
@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...

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.