# Prove for every $m, n$, $(m^2 - n^2, 2mn, m^2 + n^2)$ can make pythagorean triple

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 $$a^2 + b^2 = c^2$$?

• Have you tried simply squaring those expressions? Two of them are $a$ and $b$, and the other is $c$. – rogerl Oct 26 '18 at 0:37
• Take a look at how to use mathjax to write the mathematics in your post. – Aaron Zolotor Oct 26 '18 at 0:37
• 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? – biotecher Oct 26 '18 at 0:47
• Yes. I think the more interesting problem is showing that this generates all possible triples. – Don Thousand Oct 26 '18 at 1:03

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.,

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