On Fermat's Last Theorem, Continued (Part 2): Pythagorean Triples

从费马大定理谈起（二）：勾股数 苏剑林 · 2014-08-15 · Mathematics number theory Diophantine equation fermat's last theorem

Fermat's Last Theorem concerns the case $n > 2$, but we can start from $n=2$ and work out the general expression for Pythagorean triples, thereby getting a taste of the original idea behind the proof of Fermat's Last Theorem.

Coprime Solutions

We work within the real integers, that is, within $\mathbb{Z}$. To solve the indeterminate equation $x^2+y^2=z^2$, we first note that this is a homogeneous equation, which tells us that if a solution exists, we can always divide through by the greatest common divisor to obtain a solution in which the values are pairwise coprime. In other words, whenever a solution exists, a coprime solution exists — this is a general feature of solutions to $x^n+y^n=z^n$. So let's assume that $(x,y,z)=(a,b,c)$ is a coprime solution of the equation $x^2+y^2=z^2$. more

Let's now analyze the congruence properties. We choose the prime factor 2, that is, we examine the parity of $a,b,c$. Within $\mathbb{Z}$, 2 is a rather special prime, since it is the prime of smallest absolute value, and so it has the simplest possible congruence behavior (dividing any integer by 2 leaves either remainder 1, or no remainder at all). It's easy to notice the following fact:

If $n\equiv 1(\bmod\,2)$, then $n^2\equiv 1(\bmod\,4)$.

From this we can deduce that among $a,b$, one must be even and the other odd. Here's the proof: since we assumed a coprime solution, they cannot both be even; and if both were odd, we would have $a^2\equiv 1(\bmod\,4),b^2\equiv 1(\bmod\,4)$, giving $(a^2+b^2)\equiv 2(\bmod\,4)$, but $a^2+b^2=c^2$, so $c^2\equiv 2(\bmod\,4)$, which is a contradiction, since a perfect square mod 4 can only be 1 or 0.

Since $a,b$ consists of one odd and one even number, $c$ must be odd, so let's take $a$ to be even and $b$ to be odd. Notice that

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

We know that any common divisor of $m,n$ must also divide $m+n$ and $m-n$, and therefore, any common divisor of $c-b$ and $c+b$ must also divide $2c$ and $2b$. Since $c,b$ are both odd, $c-b$ and $c+b$ share the common divisor 2. So we may write $c-b=2u,c+b=2v,a=2\alpha$, giving us

$$\alpha^2=uv$$

At this point $u+v=c,v-u=b$, and since $b,c$ are coprime by assumption, $(u,v)=1$. When the product of two coprime numbers is itself a perfect square, then each of these two numbers, up to a unit factor, must itself be a perfect square. It's important to note that this qualification "up to a unit factor" cannot be dropped — the units in $\mathbb{Z}$ are $\pm 1$, which tells us that 6 can be written not only as $2\times 3$, but also as $(-2)\times (-3)$. So we may set $u=\pm p^2,v=\pm q^2$, giving us the general formula for coprime Pythagorean triples:

$$a=\pm 2pq,\ b=\pm q^2\mp p^2,\ c=\pm q^2\pm p^2$$

Reviewing the Steps

Let's briefly review each step of our derivation, paying attention to which steps can similarly be applied when analyzing solutions for $n > 2$. The first step is that a solution always implies a pairwise-coprime solution — this property holds for any $n$. Second, we used a congruence argument based on the prime 2; this particular step no longer applies once we get to $n > 2$, but after extending the number field and switching to a different prime factor, an analogous step remains available. Finally, we relied on a crucial property: "if the product of two coprime numbers is a perfect square, then each number, up to a unit factor, is itself a perfect square." If we replace "square" in this statement with "cube," "fourth power," and so on, the conclusion still holds — this is connected to the unique factorization property of $\mathbb{Z}$ (the fundamental theorem of arithmetic).

We've also seen that in order to solve $x^2+y^2=z^2$, we needed to factor it as $x^2=(z-y)(z+y)$, and for the sake of convenience in the analysis, we wanted to factor it into a product of linear polynomials. However, this cannot be done within the real numbers when it comes to $n > 2$; to make it possible, we need to extend the integers into the complex numbers, and this is where we begin to see the necessity of extending the number field. That said, many properties that hold quite obviously for the real integers — such as unique factorization — no longer hold in these extended "integers." This requires further techniques to overcome, and these topics will be discussed in later posts.