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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 $q$.

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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.

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  • $\begingroup$ Why is it important that $\gcd(p,q)=1$ and different parity? $\endgroup$ – Wesley Strik Oct 10 '18 at 23:45
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    $\begingroup$ @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. $\endgroup$ – Michael Rozenberg Oct 10 '18 at 23:48
  • $\begingroup$ The argument doesn't ask me to generate ALL of them, it just asks me to generate some. We really aren't that far in the subject matter yet and they have only just been introduced in an exercise. $\endgroup$ – Wesley Strik Oct 10 '18 at 23:53
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    $\begingroup$ To the proposer: If $p,q$ are of equal parity then $p^2-q^2, 2pq, $ and $p^2+q^2$ are all even. $\endgroup$ – DanielWainfleet Oct 11 '18 at 0:09
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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 $p$.

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  • $\begingroup$ I was also thrown off by that part, I mean this argument makes a whole lot more sense, but why plug in the rationals in the first place... $\endgroup$ – Wesley Strik Oct 10 '18 at 23:44

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