Why does the primitive hypotenuse of a Pythagorean triangle always be in the form $4n+1$? My understanding so far is that $z^2 = x^2 + y^2$ so based on Euler's proof $z^2 = (a^2+b^2)$ therefore $z$ is a sum of two squares (STS) and take the from $4n+1$. However, my question is, a STS can also be made up of two NON-STS because the FTA states that an even number of $4n+3$, when multiplied together, will also result in STS (eg 7*7 =49 so the primality argument falls through as well). Any thoughts is appreciated. Thank you.
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@coffeemath would you have an insight to this? – LucasCK Mar 25 '16 at 15:12
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$6, 8, 10$ is a pythagorean triple, none of those numbers are of the form $4n+1$. – Henrik supports the community Mar 25 '16 at 15:18
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@Henrik It only holds for primitve Pythagorean triples. – Dietrich Burde Mar 25 '16 at 15:19
2 Answers
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You proably want to restrict your statement to primitive pythagorean triples.
Hint: In a primitive pythagorean triple $(a, b, c)$, either $a$ or $b$ is even and the other one odd.
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It doesn't, unless you're talking about primitive triples. A counter example would be $z = 20$, $x=16$, $y=12$.
It turns out however, that primitive triples can be parametrised as $z=(u^2+v^2)$, $x=u^2-v^2$, $y=2uv$, where $\gcd(u,v)=1$ and one of $u,v$ is even and the other one is odd. If we say $u$ is odd and $v$ is even, then $u=2r+1$ and $v=2s$, for some $r,s$, so $z = 4(r^2+r+s^2)+1$, which is of the desired form.
Étienne Bézout
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Reference (http://mathworld.wolfram.com/PythagoreanTriple.html) for the formulas given by @Étienne Bézout – Jean Marie Mar 25 '16 at 15:30
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exactly the argument I was looking for! Excellent deduction, many thanks for the help. Cheers – LucasCK Mar 25 '16 at 15:41