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How can one prove that all the Pythagorean triples satisfying this condition have been found? We are working with positive integers a, b, and c.

grayQuant
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    All the Pythagorean triples can be generated by a formula. That makes it very easy to generate all the triples smaller than some given number. – J.R. Feb 16 '14 at 18:43

1 Answers1

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All Pythagorean triples can be found as $$(p^2-q^2)r,2pqr,(p^2+q^2)r$$ Consider different cases for $r=1,5,13$ and for each one consider different pairs of relatively prime $(p,q)$, where $p>q$, $p^2+q^2=65/r$.

Vadim
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  • Is this enough for proof? – grayQuant Feb 17 '14 at 00:26
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    Yes, because every Pythagorean triple can be represented as above for some $p$ and $q$. – Vadim Feb 17 '14 at 02:44
  • Can you show a few examples of how to systematically do this, maybe with c=25? – grayQuant Feb 19 '14 at 22:59
  • $c=25$, possible values for $r=1,5$, corresponding equations $p^2+q^2=25$ and $p^2+q^2=5$, first gives you $(p,q,r)=(4,3,1)$, second: $(2,1,5)$, overall for $c=25$ we have Pythagorean triples $(7,24,25),(15,20,25)$. – Vadim Feb 19 '14 at 23:25
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    Note, that this method still requires you to solve equations like $p^2+q^2=a$, but instead of searching for all solutions when $a=c^2$, you need to search for maximum $a=c$, and in this case, given that $q<p$ you search through values for $q$ less than $\sqrt{c/2}$ which is significantly lower than $\sqrt{c^2/2}=c/\sqrt{2}$. – Vadim Feb 19 '14 at 23:30