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.
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5$\begingroup$ 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. $\endgroup$– J.R.Feb 16, 2014 at 18:43
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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 $p>q$, $p^2+q^2=65/r$.
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2$\begingroup$ Yes, because every Pythagorean triple can be represented as above for some $p$ and $q$. $\endgroup$– VadimFeb 17, 2014 at 2:44
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$\begingroup$ Can you show a few examples of how to systematically do this, maybe with c=25? $\endgroup$ Feb 19, 2014 at 22:59
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$\begingroup$ $c=25$, 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 $(7,24,25),(15,20,25)$. $\endgroup$– VadimFeb 19, 2014 at 23:25
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1$\begingroup$ 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 $\sqrt{c^2/2}=c/\sqrt{2}$. $\endgroup$– VadimFeb 19, 2014 at 23:30