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April 15th, 2017, 09:16 AM   #1
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one of these that JohnG is so good at

show that $\forall a,b \in \mathbb{N}$

$m = \dfrac{a^2 + b^2}{ab+1} \in \mathbb{N} \Rightarrow \sqrt{m}\in \mathbb{N}$
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April 16th, 2017, 02:16 AM   #2
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JohnG? I think numberphile showcased this problem.
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April 16th, 2017, 04:09 AM   #3
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Let $n = \sqrt{m}$. If $n$ > 1, must $n^3$ divide $a$ or $b$, but not both?
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April 16th, 2017, 09:52 AM   #4
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Quote:
Originally Posted by romsek View Post
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.
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April 16th, 2017, 01:11 PM   #5
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There are more details here.
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April 16th, 2017, 07:31 PM   #6
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The one Terrence Tao couldn't get.
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April 17th, 2017, 08:40 AM   #7
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JohnG is good at easy number theory problems. I'm glad I don't have to think about this one.
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