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.

Advertisements
Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: