Primitive Pythagorean Triples, or PPT’s, are non-trivial integer solutions to the Pythagorean equation z² = x² + y² where we assume that 0<x<y<z, and that x, y and z are co-prime.

Euclid’s formulae allow PPT’s to be generated using 2 (positive integer) parameters, usually denoted by m and n. The PPT’s (usually) take the form ( m²– n², 2mn, m²+n²)

or ( 2mn, m²-n², m²+n²), where m>n≥1, m and n are co-prime, and m-n is odd. Clearly, m²+n² is the largest number in each triple.

Note: It is easier to spot patterns in generated triples if these are ordered as in (x, y, z) where x<y<z.

There are 2 cases that can occur, i.e. ( m²– n², 2mn, m²+n²) where m²-n²< 2mn and

( 2mn, m²-n², m²+n²) where 2mn < m²-n². If m²-n²< 2mn then it is easy to show that

n < m < (1+√2)n. Hence if we wished to write a program to generate such triples we would first have to choose n from 1,2,3, … and then choose m from (n+1), …, int((1+√2)n).

Any proofs that I have seen that establish the above formulae for x, y and z make use of parity, factorisability and co-primeness, but interestingly do not make use of geometry.

It cannot be assumed of course that x, y, and z form the sides of a triangle. That has to be proved.

However the consequences of proving that x, y, and z form the sides of a triangle lead to a simple method of generating PPT’s which is also simply proved.

New Method

Returning to the equation z²= x²+ y², it is easy to show that z<x+y, since x²+ y² < (x+y)² and so, since 0<x<y<z, x, y, and z do form the sides of a triangle. It follows, from z<x+y and x<y, that z<2y. Hence y<z<2y, and so there is an integer a, where 0<a<y, such that z=y+a. Also, since 0<x<y, there is an integer b, where 0<b<y, such that x=y-b.

So, z²= x²+ y² becomes (y+a)² = (y-b)²+ y², where 0<a<y and 0<b<y.

After multiplying out and re-arranging (y+a)² = (y-b)²+ y² we get, 2a(a+b) = (y – (a+b))².

Hence 2|(y-(a+b)) and so y-(a+b) = 2c, where c is an integer and it is easy to show that c>0.

Note: [2c = y – (a+b) = y – (z-y + y-x) = y + x – z > 0]

So, x = a+2c, y = a + b + 2c = (x+b) and z = 2a + b + 2c = (y+a), where a(a+b)= 2c².

Note: it is easy to show that a<√2 c.

To generate PPT’s, first choose any integral c, c≥1;  then choose integral a in the range 1≤a<√2c so that a|2c²; then finally choose integral b , b≥1, using a(a+b)= 2c²,

or b=(2c²)/a – a, making sure that the resulting x, y and z have no common factors.

For example:

If c=1 then a=1 and b = 1. Hence we get the PPT (3,4,5).

If c=2 then a=1 and b=7, and this gives the PPT  (5,12,13)

If c=3 then a=1 and b=17 or a=2 and b=7, giving PPT’s (7,24,25) and (8,15,17).

If c=4 then a=1 and b=31 giving PPT (9,40,41).

If c=5 then a=1 and b=49 or a=2 and b= 23 giving PPT’s ( 11, 60, 61) and ( 12, 35, 37).

If c=6 then a = 1 and b = 71 and we get the PPT (13,84,85).

Etc.

The following is a short Python program to generate PPT’s using this method.

import math
#Generating primitive Pythagorean Triples
#
def hcf(x,y):
    #assume y!=x
    while y!=x:
        if y>x:
            y=y-x
        else:
            x=x-y
        #endif
    #endwhile
    return x
#enddef
#
def generate_triples():
    for c in range(1,60):
        for a in range(1,int(math.sqrt(2)*c)+1):
            if int(2*c*c/a)==2*c*c/a:
                b=int(2*c*c/a)-a
                x=a+2*c
                y=a+b+2*c
                z=2*a+b+2*c
                if hcf(x,y)==1:
                    print ("c = ",c)
                    print ("x = ",x)
                    print ("y = ",y)
                    print ("z = ",z)
                    print ()
                #endif
            #endif
        #endfor
    #endfor
#enddef
#
generate_triples()