Exercise 5.3.1 (First Pythagorean triples)

Proof. By Theorem 5.5, the positive primitive solutions of $x^2+y^2=z^2$ with $y$ even are

where $r,s$ are arbitrary integers of opposite parity with $r>s>0$ and $r\wedge s=1$.

If $0<z<30$, then $30>z=r^2+s^2>r^2$, thus $r\le \lfloor \sqrt{30}\rfloor =5$, so

We obtain all the possibilities with this Python program:

from math import gcd, floor, sqrt
max_z = 30
max_r = floor(sqrt(max_z))
for r in range(1, max_r + 1):
    for s in range(1, r):
        if gcd(r,s) == 1 and r % 2 != s % 2:
            if r * r + s * s < max_z:
                t = [r * r - s * s, 2 * r * s, r * r + s * s]
                print((r,s), ’=>’, t)

(2, 1) => [3, 4, 5]
(3, 2) => [5, 12, 13]
(4, 1) => [15, 8, 17]
(4, 3) => [7, 24, 25]
(5, 2) => [21, 20, 29]

The positive primitive solutions of $x^2+y^2=z^2$ with $y$ even and $0<z<30$ are

and all positive primitive solutions are

(So all primitive Pythagorean triples for which $0<z<30$ are