# 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 $$q$$.

It should be $$(p^2-q^2)^2+(2pq)^2=(p^2+q^2)^2,$$ where $$p$$ and $$q$$ are natural numbers such that $$p>q$$, $$\gcd(p,q)=1$$ and $$p$$ and $$q$$ have different parity.
Now, you can get all triples: $$(d(p^2-q^2),d(2pq),d(p^2+q^2)),$$ where $$d$$ is a natural number.
• Why is it important that $\gcd(p,q)=1$ and different parity? – Wesley Strik Oct 10 '18 at 23:45
• @WesleyGroupshaveFeelingsToo If you want to get all triples then you need to assume before that $\gcd(a,b,c)=1$ in the equation $a^2+b^2=c^2$. All these assumptions follow from the proof. – Michael Rozenberg Oct 10 '18 at 23:48
• To the proposer: If $p,q$ are of equal parity then $p^2-q^2, 2pq,$ and $p^2+q^2$ are all even. – DanielWainfleet Oct 11 '18 at 0:09
I don't see why you want to substitute in rationals. If you substitute in whole numbers for $$X$$ you get a Pythagorean triple $$X^2-1,2X,X^2+1$$. That is three numbers just like you are looking for. If you multiply your last by $$q^2$$ you clear the fractions and get the same triple based on $$p$$.