Problem 5

Three positive integers $a$, $b$, and $c$ form a Pythagorean Triple if $a^2+b^2=c^2$. It is primitive if they have no common factors. For example, $3$, $4$ and $5$ form a primitive Pythagorean Triple, $5$, $12$, and $13$ form another primitive Pythagorean Triple. Find two more primitive Pythagorean Triples.

Commentary: I had already read about how the set of Pythagorean Triples may be completely classified, though I didn't quite recall it at the moment, so I decided to do this problem by figuring out my own simple method.

Proof: Observe $a^2+b^2=c^2$ implies $a^2=(c-b)(c+b)$, so when $a$ is an odd prime this means we must solve $c-b=1$ and $c+b=a^2$, namely by $c=b+1$ and $b=(a^2-1)/2$. The triples $(3,4,5)$ and $(5,12,13)$ are of this form, and we may generate $(7,24,25)$, $(11,60,61)$, $...$.$~\square$

Since $(a,b,c)$ being a triple implies $(an,bn,cn)$ is a triple for $n∈ℕ$, this implies that every positive integer with an odd prime divisor (i.e., every positive integer except powers of $2$) is part of a Pythagorean Triple. This method can be applied to demonstrate that $1$ and $2$ are not part of any Pythagorean Triple, and that $(2^n,(2^{2n-1}-2)/2,(2^{2n-1}+2)/2)$ for $n > 1$ is a nontrivial Pythagorean Triple for greater powers of $2$.