Time to finish what we started last week.

We saw that if *a, b, c* was a primitive Pythagorean triple, then at least one of *a* and *b* is even and one is odd. Let us declare, then, that we will use *a* to denote the odd length and *b* to denote the even one. By rearranging the Pythagorean equation and factoring we get:

\[

a^2=c^2-b^2=(c+b)(c-b).

\]

Let’s try this out for a few specific triples:

\[

3^2=(5+4)(5-4)=(9)(1)

\]

\[

5^2=(13+12)(13-12)=(25)(1)

\]

\[

15^2=(17+8)(17-8)=(25)(9)

\]

\[

45^2=(53+28)(53-28)=(81)(25)

\]

In each case we find that *c+b* and *c-b* are both perfect squares. Even more important, they are relatively prime. Let’s see if we can prove that.

We begin by noting that, since *b* is even and *c* is odd, we have that *c+b* and *c-b* are both odd.

Now suppose that *d* is some number that divides both of them. We want to show that *d* has to be one. Since *d* divides both *c+b* and *c-b*, we must also have that *d* divides their sum and their difference. That is, *d* divides:

\[

(c+b)+(c-b)=2c \phantom{xxx} \textrm{and} \phantom{xxx} (c+b)-(c-b)=2b

\]

Recall, though, that we are discussing **primitive** Pythagorean triples. That means that *b* and *c* are relatively prime. It follows that *d* must divide *2*, meaning that *d=1* or *d=2*.

Having already established that *c+b* and *c-b* are both odd, we know that *d* is not two. We conclude that *d=1* as desired.

This is significant for the following reason: If the product of two relatively prime numbers is a perfect square, then the numbers must themselves be perfect squares. I won’t stop to prove that here, though it is straightforward if you think in terms of prime factorizations.

We can now write

\[

c+b=s^2 \phantom{xxx} \textrm{and} \phantom{xxx} c-b=t^2,

\]

where *s* and *t* are odd, relatively prime integers. But now we can solve for *b* and *c* to obtain

\[

c= \frac{s^2+t^2}{2} \phantom{xxxx} \textrm{and} \phantom{xxxx} b=\frac{s^2-t^2}{2},

\]

from which it follows that

\[

a=st.

\]

And we’re done! So long as *s* and *t* are odd and relatively prime, these formulas will give you a primitive Pythagorean triple. Moreover, all such triples can be produced in this way. For example, choosing *s=3* and *t=1* gives you *(3,4,5)* and choosing *s=5* and *t=1* gives you *(5,12,13)*.

As was pointed out in the comments to last week’s post, we could also take a purely geometric approach to this problem. Notice that if *(a,b,c)* is a Pythagorean triple, then we have

\[

\left( \frac{a}{c} \right)^2 + \left( \frac{b}{c} \right)^2 =1,

\]

meaning that any such triple corresponds to a rational point on the curve

\[

x^2+y^2=1

\]

That’s the equation for the unit circle. Imagine now that we draw a line, with rational slope *m*, through the point *(-1,0)*. If you remember your elementary algebra you will see that the equation for this line is

\[

y=m(x+1)

\]

This line will hit the unit circle in exactly one other point, as shown below:

It is now a straightforward algebraic exercise to solve our equations simultaneously to work out the coordinates of the second point of intersection, which will be a rational point on the circle. When that is done, you quickly recover our formulas for producing Pythagorean triples. Very nice!