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 April 15th, 2017, 09:16 AM #1 Senior Member     Joined: Sep 2015 From: USA Posts: 2,125 Thanks: 1103 one of these that JohnG is so good at show that $\forall a,b \in \mathbb{N}$ $m = \dfrac{a^2 + b^2}{ab+1} \in \mathbb{N} \Rightarrow \sqrt{m}\in \mathbb{N}$
 April 16th, 2017, 02:16 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,722 Thanks: 600 Math Focus: Yet to find out. JohnG? I think numberphile showcased this problem.
 April 16th, 2017, 04:09 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,731 Thanks: 1808 Let $n = \sqrt{m}$. If $n$ > 1, must $n^3$ divide $a$ or $b$, but not both?
April 16th, 2017, 09:52 AM   #4
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Joined: Sep 2015
From: USA

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Thanks: 1103

Quote:
 Originally Posted by romsek show that $\forall a,b \in \mathbb{N}$ $m = \dfrac{a^2 + b^2}{ab+1} \in \mathbb{N} \Rightarrow \sqrt{m}\in \mathbb{N}$
I found an answer. A very common problem apparently.

Last edited by romsek; April 16th, 2017 at 09:55 AM.

 April 16th, 2017, 01:11 PM #5 Global Moderator   Joined: Dec 2006 Posts: 19,731 Thanks: 1808 There are more details here.
 April 16th, 2017, 07:31 PM #6 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,722 Thanks: 600 Math Focus: Yet to find out. The one Terrence Tao couldn't get.
 April 17th, 2017, 08:40 AM #7 Member   Joined: Jan 2016 From: Athens, OH Posts: 89 Thanks: 47 JohnG is good at easy number theory problems. I'm glad I don't have to think about this one.

 Tags good, johng

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