So we have \(\gcd(x,y,z)=1\) and \(x,z\) are odd and \(y\) is even. Now we will do a somewhat intricate, but familiar, type of argument about factorization and divisibility.

Let's rewrite our situation as \begin{equation*}y^2=z^2-x^2\; .\end{equation*}The right-hand side factors as \begin{equation*}z^2-x^2=(z-x)(z+x)\, .\end{equation*} Certainly \(z-x\) and \(z+x\) are both even, so that \(z-x=2m\) and \(z+x=2n\). But since their product is a square (\(y^2\)), then that product \(2m\cdot 2n=4mn\) is also a perfect square. Since \(y\) is even, \(y=2k\) for some \(k\in\mathbb{Z}\) and \(y^2=4k^2\), so \(mn=k^2\) is a perfect square.

Let's look at these mysterious factors \(m=\frac{z-x}{2}\) and \(n=\frac{z+x}{2}\). Are they relatively prime? Well, if they shared a factor, then \(x=m+n\) and \(z=n-m\) also share that factor. But \(\gcd(x,z)=1\), so there are no such factors and \begin{equation*}\gcd\left(\frac{z-x}{2},\frac{z+x}{2}\right)=\gcd(m,n)=1\end{equation*}

Now recall that \(y\) is even. Letting \(y=2j\), we see that \(y^2=4mn\) yields \(j^2=mn\); however, this time \(m\) and \(n\) are relatively prime!

At this point we need what may seem to be an intuitive fact about squares and division; if coprime integers make a square when multiplied, then they are each a perfect square. (See Proposition 3.7.2.) So \(m=p^2\) and \(n=q^2\) for some integers (obviously coprime) \(p\) and \(q\).

This clearly implies that \(j^2=p^2q^2\), so \(y=2pq\).

Now we can put everything together. \begin{equation*}z-x=2p^2,\; z+x=2q^2,\; \text{ and }y=2pq\, . \end{equation*} That is:

It's really worth trying to find these by hand; it gives one a very good sense of how this all works.

Of course, you could generate some by computer as well …

Historically, one of the big questions one could ask about such Pythagorean integer triangles was about its area. For primitive ones, the legs must have opposite parity (do you remember why?), so the areas will be integers. (For ones which are not primitive, the sides are multiples of sides with opposite parity, so they are certainly also going to have an integer area.)

So what integers work? You all know one with area 6, and it should be clear that ones with area 1 and 2 can't work (because the sides would be too small and because \(2,1\) doesn't lead to a triple); can you find ones with other areas?

It is worth asking why there are no odd numbers in the list so far. In fact, we can prove quite a bit about these things.

Remember, \(x\) and \(y\) can be written as \(x=q^2-p^2\) while \(y=2pq\), for relatively prime opposite parity \(q>p\). Then the area must be \begin{equation*}pq(q^2-p^2)=pq(q+p)(q-p)\, .\end{equation*} So can the area be odd?

So one could analyze a number to see if it is possible to write as a product of four relatively prime integers as a starting point. E.g. \(30=2\cdot 3\cdot 5\cdot 1\) is the only way to write \(30\) as a product of four such numbers (assuming no more than one of those is 1!), and since \(q+p\) must be the biggest, we must set \(q+p=5\). Quickly one can see that \(q=3,p=2\) works with this, so there is such a triangle. What are the sides?

This turns out to be a very deep unsolved problem. This news update from the American Institute of Mathematics gives some background on the congruent number problem, which asks the related question of which Pythagorean triangles with rational side lengths give integer areas. This page in particular is interesting from our present point of view.

But we can ask another question, which led Fermat to some of his initial investigations into this theory.

If you'll notice, we don't see to be getting a lot of these. In fact, none. What would we need to do to investigate this?

Remember, \(x\) and \(y\) can be written as \(x=q^2-p^2\) while \(y=2pq\), for relatively prime opposite parity \(q>p\). Then the area must be \begin{equation*}pq(q^2-p^2)=pq(q+p)(q-p)\, .\end{equation*} Now, in the previous section, we showed each of these quantities was relatively prime to each other. So if the area is also a perfect square, then since they are coprime, they themselves are all perfect squares!

Now we will do something very clever. It is a proof strategy, similar to something the Greeks used occasionally, and it is something Fermat used for many of his proofs, called infinite descent. We are going to take that (hypothetical) triangle, and produce a triangle with strictly smaller sides but otherwise with the same properties – including integer sides and square area! That means we could apply the same argument to our new triangle, and then the next one … but the Well-Ordering Principle (1.2.1) won't allow that. So the original triangle was impossible to begin with.

So let's make that smaller triangle!

In Exercise 3.6.9 you will use this to prove the famous first case of Fermat's Last Theorem: \begin{equation*}x^4+y^4=z^4\end{equation*} is not possible for any three positive integers \(x,y,z\). See also Subsection 14.2.2.