It just so happens that I am teaching elementary number theory this term. So how about the triumphant return of Monday Math!

For those playing the home game, the course textbook is *A Friendly Introduction to Number Theory (Third Ed.)* by Joseph Silverman. Let’s begin.

I’m sure we all remember the Pythagorean Theorem. That’s the one that says that the sides of a right triangle satisfy the equation:

\[

a^2+b^2=c^2

\]

where *c* is the length of the hypotenuse (the side opposite the right angle). You might wonder if it is possible for *a*, *b* and *c* to all be integers. Indeed it is, as the following equation shows:

\[

3^2+4^2=5^2

\]

The next question, then, is whether there are infinitely many such “Pythagorean triples,” that is, triples of integers that satisfy the Pythagorean equation. Yes there are, as the following list suggests:

\[

\begin{matrix}

(6, 8, 10) & (9, 12, 15) \\

(12, 16, 20) & (15, 20, 25)

\end{matrix}

\]

Of course, we obtained this list by multiplying *(3, 4, 5)* by two, three, four and five. It is not hard to prove that if we multiply the entries of a Pythagorean triple by the same number, the result is another triple.

Somehow, though, that doesn’t really seem to be in the spirit of things. It seems reasonable to say that the four triples above, and all the other ones I could obtain in a similar fashion, are effectively identical to *(3,4,5)*. When we ask for infinitely many Pythagorean triples, we are really asking if there are infinitely many *fundamentally different* triples. That leads us to a definition.

We shall say a Pythagorean triple is **primitive** if the three entries have no common factor (other than one, of course). Thus, *(3, 4, 5)* is primitive, as is *(5, 12, 13)*. But *(21, 28, 35)* is not primitive since all of the entries are multiples of seven.

It turns out that there are, indeed, infinitely many primitive Pythagorean triples, but the proof of that requires a few steps. Let’s get started today, and we will see the rest in a future post.

Here’s a list of all the primitive Pythagorean triples with hypotenuse smaller than 100:

\[

\begin{matrix}

(3, 4, 5) & (5, 12, 13) & (7, 24, 25) & (8, 15, 17) \\

(9, 40, 41) & (11, 60, 61) & (12, 35, 37) & (13, 84, 85) \\

(16, 63, 65) & (20, 21, 29) & (28, 45, 53) & (33, 56, 65) \\

(36, 77, 85) & (39, 80, 89) & (48, 55, 73) & (65, 72, 97)

\end{matrix}

\]

Notice anything? It seems that in every triple the hypotenuse is odd, and the legs are of different parity (that is, one is even and one is odd). That is not a coincidence.

It is easy to see that we can’t have that *a* and *b* are both even. For if they were it would follow immediately, since the sum of two even numbers is even, that *c* must be even as well. That would mean that *a, b, c* are all even, which is impossible in a primitive Pythagorean triple.

What happens if *a* and *b* are both odd? In this case, since the sum of two odd numbers is even, we would have that *c* is even. Therefore, we can write, for some appropriate values of *x*, *y* and *z* that:

\[

a=2x+1 \phantom{xxxxx} b=2y+1 \phantom{xxxxx} c=2z.

\]

Plugging into the Pythagorean equation gives:

\[

\left( 2x+1 \right)^2 + \left( 2y+1 \right)^2 = \left(2z)^2 ,

\]

and multiplying out now gives:

\[

\left(4x^2+4x+4y^2+4y \right)+2=4z^2

\]

This, alas, is simply impossible. The right-hand side is a multiple of four. But the left-hand side is plainly two more than a multiple of four. We conclude that we can not have that *a* and *b* are both odd. The only remaining possibility is that one of the legs is odd and the other is even. This forces the hypotenuse to be odd.

Good progress, but we still have some work to do. Let’s save that for next week!