We consider the diophantine equation
……………………………………………………. (1)
whose connection to Pythagoras’ Theorem is obvious.
In what follows we take (a_{1}, a_{2}, … a_{n}) to be the greatest common divisor of the integers in the brackets.
It is clear that any positive solution a, b, c determines an infinite number of solutions ± ta, ± tb, ± tc where t is any integer, and also that 0, 0, 0 and ±a, 0, ±a and 0, ±a, ±a are solutions so we can confine our attention to cases where (x, y, z) = 1 and x, y, z > 0. We call these solutions the Pythagorean Numbers.
Proposition: the complete set of Pythagorean numbers without duplicates is:
; ;
where a, b run through all the positive integers such that
a > b > 0 ; (a, b) = 1 and one of a and b is even, one odd.
Note: The first few such numbers are:
a 
b 
x 
y 
z 

2 
1 
3 
4 
5 

3 
2 
5 
12 
13 

4 
1 
15 
8 
17 

4 
3 
7 
24 
25 

Proof: We take as given the following Lemma: if (m, n) = 1 and p divides mn then either p divides m or p divides n. For a proof see ‘Common Factors and Linear Diophantine Equations’ accessible through ‘Notes on Numbers’ (see sidebar on the home page).
First we show that if the triplets x, y, z constructed as per the proposition are Pythagorean numbers. It is clear by substitution that they are solutions of (1). It is also clear that they are all positive. Suppose n > 1 is some common divisor of x, y and z. z = (a + b)^{2} – 4ab is odd so n is odd. Since n divides y =2ab, it must divide either a or b, but it cannot divide both. Thus n divides either a^{2} or b^{2} but cannot divide both. Thus n cannot divide z, a contradiction. Thus (x, y, z) = 1.
The set of triplets so constructed contains no duplicates since if y = 2a_{1}b_{1} = 2a_{2}b_{2} then by the lemma and the Unique Factorisation Theorem a_{1} = a_{2} and b_{1} = b_{2}.
Now we show that if x, y, z is a Pythagorean triplet then it either has the form given in the Proposition, or is a duplicate
(x,y) = 1 because if n > 1 divides x and y it divides z, a contradiction. Thus x,y cannot both be even. If x is odd then x^{2} leaves remainder 1 on division by 4, and similarly with y, so x,y cannot both be odd. Therefore z must be odd.
Suppose x is odd. We can factor (1) as y^{2} = z^{2} – x^{2} = (z + x)(z – x). Clearly 2 divides both z + x and z – x. Suppose there is a larger common divisor n. Then n divides (z + x) + (z – x) = 2z and (z + x) – (z – x) = 2x, so n/2 divides both z and x, and so (n/2)^{2} divides z^{2} – x^{2} = y^{2} and (x, y, z) = n/2 > 1, a contradiction.
Consequently we can write (y/2)^{2} = ((z + x)/2)((z – x)/2) where ((z + x)/2,(z – x)/2) = 1. Then as a consequence of the Unique Factorisation Theorem, (z + x)/2 and (z – x)/2 must both be squares.
So we can write (z + x)/2 = a^{2}, (z – x)/2 = b^{2}, (a,b) = 1. Therefore z = a^{2} + b^{2}, x = a^{2} – b^{2}, y = 2ab. Since x > 0, a > b and since x,z are both odd, one of a and b must be even.
Likewise, if we suppose y is odd we can factor (1) as x^{2} = z^{2} – y^{2} = (z + y)(z – y) and show that the triplet x, y, z has the duplicate form given above.
A pdf version of this post can be found here.
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