<s>[INST] Generating Pythagorean triples I'm asked to generate Pythagorean triples from the polynomial identity:
$$(X^2-1)^2 + (2X)^2=(X^2+1)^2$$
By substituting rational numbers $\frac p q$ for $X$. However, Pythagorean triples are just as the name says, it, three numbers. If I would substitute this number I get:
$$(\left(\frac p q\right)^2-1)^2 + 4\left(\frac p q\right)^2=(\left(\frac p q\right)^2+1)^2$$
How would I get three integers from this? There are just two numbers involved, $p$ and $q$.
 [/INST] If we take the equation asis, it is not true unless $x=0$ or$X=1$.
$$(x-1)^2+2x=(x+1)^2$$
$$(x^2-2X+1)+4X^2=(x^2+2X+1)$$
Subracting $x^2+2x+1$ from both sides we get
$$4X^2-4X=0\Rightarrow4x(X-1)=0$$
A variation of the identity will work for all $n$ in generating triples where $C-A)=2.$
$$\mathbf{A=4n^2-1}$$
$$\mathbf{B=4n}$$
$$\mathbf{C=4n^2+1}$$
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