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I tried researching more about this because it seems to be a common topic, but I don't know how to approach this problem. Do I have to somehow arrange those 3 terms into $a^2 + b^2 = c^2$?

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  • $\begingroup$ Have you tried simply squaring those expressions? Two of them are $a$ and $b$, and the other is $c$. $\endgroup$ – rogerl Oct 26 '18 at 0:37
  • $\begingroup$ Take a look at how to use mathjax to write the mathematics in your post. $\endgroup$ – Aaron Zolotor Oct 26 '18 at 0:37
  • $\begingroup$ So then for the proof am I supposed to add the squares of the first two terms and show it reduces into the square of the third term? $\endgroup$ – biotecher Oct 26 '18 at 0:47
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    $\begingroup$ Yes. I think the more interesting problem is showing that this generates all possible triples. $\endgroup$ – Don Thousand Oct 26 '18 at 1:03
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Not too complicated:

$(m^2 - n^2)^2 + (2mn)^2 = m^4 - 2m^2 n^2 + n^4 + 4m^2n^2$ $= m^4 + 2m^2n^2 + n^4 = (m^2 + n^2)^2. \tag 1$

Now if $N$ is odd set

$N = 2s + 1; \tag 2$

then we take

$m = s + 1, \; n = s; \tag 3$

if $N$ is even, write

$N = 2s, \tag 4$

and

$2mn = 2s \Longrightarrow mn = s, \tag 5$

e.g.,

$m = s, \; n = 1. \tag 6$

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