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Show that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet

First, I showed that $(m^2 - n^2, 2mn, m^2 + n^2)$ is in fact a Pythagorean triplet.

$$\begin{align*} (m^2 - n^2)^2 + (2mn)^2 &= (m^2 + n^2)^2 \\ &= m^4 -2m^2n^2 + n^4 + 4m^2n^2 \\ &= m^4 + 2m^2n^2 + n^4 \\ &= 1\end{align*}$$ which shows that it respect $a^2+b^2 = c^2$

let p be a prime number, $ p|(m^2 + n^2) \text { and } p|(m^2 - n^2) $

if $gcd(m^2 + n^2, (m^2 - n^2)) = 1$

$p | (m^2 + n^2) , \text { so, } p |m^2 \text { and } p |n^2$

that means $ (m^2 + n^2) \text { and } (m^2 - n^2) $ are prime together

I'm kind of lost when I begin to show the gcd = 1... I think I know what to do, just not sure how to do it correctly.

Thanks

Cortizol
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Dave
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    One nitpick; when proving an equality like you are asked to do, you should start from one side of the equality and work your way to other side of the equality. Written as you have seems to 'assume' the equality from the beginning and works your way down to something you know is true; proofs go in the opposite direction. – Clayton Feb 17 '13 at 18:51
  • You have an extraneous $==1$ at the end – Ross Millikan Feb 20 '13 at 15:59

3 Answers3

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You don't know that $\gcd(m^2-n^2, m^2+n^2)=1$, you need to prove it.

Now if $p|m^2-n^2$ and $p|m^2+n^2$ then $p$ divides their sum and their difference. Use this.

And to proceed further you need extra information, which you probably left out but it is the key: $m,n$ are relatively prime and of opposite parity....

N. S.
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To show $(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2$ is equivalent to showing $(m^2 - n^2)^2 + (2mn)^2 - (m^2 + n^2)^2 = 0$ so \begin{align*} && (m^2 - n^2)^2 + (2mn)^2 - (m^2 + n^2)^2 \\ &=& m^4 - 2m^2n^2 + n^4 + 4m^2n^2 - m^4 - 2m^2n^2 - n^4 \\ &=& m^4 + n^4 - m^4 - n^4 \\ &=& 0\end{align*}

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    Despite the OP having done the calculation incorrectly, the question was not about showing that $m^2 - n^2, 2mn, m^2 + n^2$ was a Pythagorean triple, but rather that it is primitive. – JavaMan Feb 20 '13 at 15:56
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The question is wrong. It is not always primitive. As when we put m=3, n=1 The triplet comes to be 8,6,10, which is not a primitive Pythagorean triplet.

user1059633
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