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

Question

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

Answer 1

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.,

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

Answer 2

It's hard to do with Euclid's formula but here are functions I developed to show there is a Pythagorean triple for every pair of natural numbers $n$ and $k$.

Theorem: $$\forall n,k \in \mathbb{N}, \exists A,B,C\in \mathbb{N}:A^2+B^2=C^2\text{ if } A=(2n-1)^2+2(2n-1)k$$

Proof: Let $$A=(2n-1)^2+2(2n-1)k$$

Solving $A^2+B^2=C^2$ for $B$ and $C$, respectively and substituting $A$, we find that $$B=2(2n-1)k+2 k^2$$and$$C=(2n-1)^2+2(2n-1)k+2k^2$$ We can then show that $$A^2=(2n-1)^4+4(2n-1)^3 k+4(2n-1)^2 k^2$$ $$B^2=4(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$ $$C^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4$$ $$A^2+B^2=(2n-1)^4+4(2n-1)^3 k+8(2n-1)^2 k^2+8(2n-1) k^3+4k^4=C^2$$

$\therefore \forall n,k \in \mathbb{N},\exists A,B,C\in \mathbb{N}:A^2+B^2=C^2 \text{ if } A=(2n-1)^2+2(2n-1)k\text{ } \blacksquare$