If $a@, $b@, and $c@ are all positive integers, this is called
a **Pythagorean triple**.
The best known example, and the simplest one, is $3@, $4@, $5@ since
$$ 9 + 16 = 25 @@

and another well known one is $5@, $12@, $13@.

*How can we find all Pythagorean triples?*

If $a, b, c@ is a Pythagorean triple, then
so is $ma, mb, mc@ for any positive integer
$m@. Thus in order to find all of them,
it suffices to find those which are **primitive** - i.e.
for which there does not exist a common divisor of all
three coordinates.

- If $a,b,c@ is a Pythagorean triple, then it is primitive if and only if ts coordinates are pair-wise relatively prime.

Because if a prime number $p@ divides two of them, say $a@ and $b@, it will also divide $c^{2}@, hence $c@.

If $a@ and $b@ are both even, then $a^{2}+b^{2}@ will be even as well, and they cannot be part of a primitive Pythagorean triple. If $a@ and $b@ are both odd, then $a^{2}+b^{2}@ will be congruent to $2@ modulo $4@, and it cannot be the square of an integer. Therefore in a primitive Pythagorean triple exactly one of $a@ and $b@ is odd and one even.

If $(a, b, c)@ is a primitive Pythagorean triple, then the point $(a/c, b/c)@ is a point on the unit circle with rational coordinates expressed as reduced fractions. In fact, finding all Pythagorean triples turns out to be nearly equivalent to finding all such points.

*How can we find all points
$(A, B)@ on the unit circle with rational coordinates?*

If $(A, B)@ is a rational point on the unit circle other than $(1,0)@, then the line through $(A,B)@ and $(1,0)@ has the equation $$ y/(x-1) = B/(A-1) = m @@

whose slope is the rational number $m = B/(A-1)@. Conversely, if $m@ is any rational number then the line through $(1,0)@ with slope $m@ intersects the unit circle at one other point. Thus there is exactly one rational point on the unit circle for each rational slope. Those with positive coordinates correspond to slopes $m@ less than $-1@.

B = -2m / (m^{2}+1) . @@

B = -2pq / (p^{2}+q^{2}) . @@

and if exactly one of $p@, $q@ is even and one odd
then the triple
$$
a = p^{2}-q^{2}

b = 2 pq

c = p^{2}+q^{2} .
@@

will be a primitive Pythagorean triple.
Conversely, suppose that $a@, $b@, $c@ be a primitive
Pythagorean triple.
One of $a@, $b@ is even - say $b = 2n@.
Then
$$ 4n^{2} = c^{2} - a^{2}

n^{2} = [(c-a)/2][(c+a)/2]
@@

Since $c@ and $a@ are relatively prime,
so are the factors $(c-a)/2@ and $(c+a)/2@, so that each of these
is also a square. Set
$$
p^{2} = (c+a)/2

q^{2} = (c-a)/2
@@

- O. Neugebauer,
**The exact sciences in antiquity**, Dover, 1969. This is a reprint of the 1957 second edition. In Chapter 2 is a discussion of how the Babylonians might have come across the parametrization we have described.

Written by Bill Casselman.