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# The parametrization of Pythagorean triples

In a right triangle with sides *a* and *b* and hypotenuse
*c* we have
* a*^{2} + b^{2} = c^{2} .
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*.

Explicitly, if *m* is the slope
then
*
A = (m*^{2}-1)/(m^{2}+1)

B = -2m / (m^{2}+1) .
### Rational points on the circle and primitive Pythagorean triples

If *m=-p/q* with *p Y q Y 0* and relatively prime, then
*
A = (p*^{2}-q^{2})/(p^{2}+q^{2})

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
which implies that *b = 2pq*.

## References

- 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.