## Find all the Pythagorean triples with one parameter fixed

Find all of the positive Pythagorean triples $(x,y,z)$ where $y = 60$.

First we find all of the positive primitive solutions. If $(x,60,z)$ is a primitive Pythagorean triple, then by Theorem 11.1, we have integers $r$ and $s$ such that $x = r^2 - s^2$, $60 = 2rs$, and $z = r^2 + s^2$, where $0 > s > r$, $(r,s) = (1)$, and $r+s \equiv 1$ mod 2. Now $rs = 30$; there are only four possibilities for $(r,s)$ with these constraints: $(r,s) \in \{(30,1),(15,2),(10,3),(6,5)\}$. These yield the triples $(899,60,901)$, $(221,60,229)$, $(91,60,109)$, and $(11,60,61)$.

If $(x,60,z)$ is not primitive, then $x$, $z$, and $60$ have a greatest common factor $d \neq 1$. There are 11 possibilities: $d \in \{2,3,4,5,6,10,12,15,20,30,60\}$. Then $(x/d,60/d,z/d)$ is primitive. First, we have a lemma.

Lemma: Consider the triple $T = (a,b,c)$ of positive integers. $T$ is not primitive Pythagorean if any of the following hold: (1) if $b$ is odd, (2) if $b = 2m$ where $m$ is odd, and (3) if $b = 2$. Proof: If $(a,b,c)$ is a primitive Pythagorean triple, then we have integers $r$ and $s$ such that $b = 2rs$, $0 < s < r$, $(r,s) = (1)$, and $r+s\equiv 1$ mod 2. If $b$ is odd, we contradict $b = 2rs$. If $b = 2m$ where $m$ is an odd, then $r$ and $s$ are both odd, but then $r+s \equiv 0$ mod 2. If $b = 2$, then $r = s = 1$, a contradiction. $\square$

By the lemma, no primitive Pythagorean triples $(x/d,60/d,z/d)$ exist for $d$ in $\{2,4,6,10,12,20,30,60\}$. We handle each remaining case in turn.

$(d = 3)$ Let $a = x/3$ and $c = z/3$, and suppose $(a,20,c)$ is a primitive Pythagorean triple. By Theorem 11.1, we have integers $r$ and $s$ such that $a = r^2 - s^2$, $20 = 2rs$, $c = r^2+s^2$, $0 < s < r$, $(r,s) = (1)$, and $r+s \equiv 1$ mod 2. Now $rs = 10$; there are two possibilities for $(r,s)$ with these constraints: $(r,s) \in \{(10,1),(5,2)\}$. These yield the solutions $297^2 + 60^2 = 303^2$ and $63^2 + 60^2 = 87^2$.

$(d = 5)$ Let $a = x/5$ and $c = z/5$, and suppose $(a,12,c)$ is a primitive Pythagorean triple. By Theorem 11.1, we have integers $r$ and $s$ such that $a = r^2 - s^2$, $12 = 2rs$, $c = r^2+s^2$, $0 < s < r$, $(r,s) = (1)$, and $r+s \equiv 1$ mod 2. Now $6 = rs$; the only such pair is $(r,s) = (3,2)$, which yields the solution $25^2 + 60^2 = 65^2$.

$(d = 15)$ Let $a = x/15$ and $c = z/15$, and suppose $(a,4,c)$ is a primitive Pythagorean triple. By Theorem 11.1, we have integers $r$ and $s$ such that $a = r^2 - s^2$, $4 = 2rs$, $c = r^2+s^2$, $0 < s < r$, $(r,s) = (1)$, and $r+s \equiv 1$ mod 2. Now $rs = 2$, so that $(r,s) = (2,1)$. This yields the solution $45^2 + 60^2 = 75^2$.