Can the area of a rational right triangle be an integer?

有理直角三角形的面积能否为整数？ 苏剑林 · 2011-08-21 · Mathematics prime number prime number conjecture integer rational number

This is an old and interesting question, but before introducing it, let's first look at a simpler problem:

Can the area of an integer-sided right triangle be a perfect square?

The answer is no. We can check a few examples: the right triangle with sides 3, 4, 5 has area 6, and 6 is not a perfect square; the right triangle with sides 5, 12, 13 has area 30, and 30 is not a perfect square either... Of course, the real aim of mathematics is a rigorous proof, not just a handful of examples—otherwise this would only be called incomplete induction, giving us a conjecture rather than a "theorem," much like the famous Goldbach conjecture. In this post we will try to prove this statement.

We'll also see shortly that this problem is equivalent to the following:

Does there exist a right triangle with area 1 whose three side lengths are all rational?

Even more surprisingly, this problem is also equivalent to showing that the equation $x^4+y^4=z^4$ has no integer solutions—in other words, we need to prove "Fermat's Last Theorem" for the case n = 4! more

Let's start from the original problem and work through the proof. First we need to establish three lemmas:

1. If a and b are two coprime integers of opposite parity (one odd, one even), then a, b, (a+b), (a-b) are pairwise coprime.

1. If the product of n pairwise coprime integers is a perfect square, then each of those n integers is itself a perfect square.

1. Every Pythagorean triple can be written in the form $(a^2-b^2)^2+(2ab)^2=(a^2+b^2)^2$, where a and b are positive integers.

These lemmas are not hard to prove using some basic theorems of elementary number theory, so we won't give the details here. As a hint for the second lemma: if the perfect square has a prime factor p, then the number of times p appears must be even; when we distribute these factors of p among the n integers, if they were split between two different integers, those two integers would then share the common factor p, contradicting the premise that the n integers are pairwise coprime. So all copies of p must be assigned to a single integer. Since this holds for every prime factor of the square, each of the n integers must consist of an even number of prime factors, meaning each one is itself a perfect square.

Now let's turn to integer-sided right triangles, using proof by contradiction. Suppose there exists such a triangle whose area is a perfect square. Then there must exist a right triangle with coprime legs whose area is a perfect square, and among all such triangles, take the one with the smallest perimeter (such a minimal one must exist within the set of integer-sided triangles). We know every Pythagorean triple can be written as $(a^2-b^2)+(2ab)^2=(a^2+b^2)^2$; since we've assumed the two legs are coprime, a and b are coprime integers of opposite parity, with a > b.

The area of the triangle is $1/2 \cdot 2ab \cdot (a^2-b^2)=ab(a-b)(a+b)$, which, by assumption, is a perfect square. By Lemma 1, a, b, (a+b), (a-b) are pairwise coprime, so by Lemma 2, a, b, (a+b), (a-b) are all perfect squares. Let $a=x^2,b=y^2$, and write

$$\begin{aligned}x^2+y^2=z^2 \\ x^2-y^2=w^2\end{aligned}$$

It's easy to see that y must be even.

Multiplying the two together gives $x^4-y^4=(zw)^2$, which shows that $(zw),y^2,x^2$ is also a Pythagorean triple. By Lemma 3, we may write $zw=p^2-q^2,y^2=2pq,x^2=p^2+q^2$, where p and q are also positive integers. Then there exists a right triangle with sides p, q, x whose area is $1/2 pq=(\frac{y}{2})^2$, satisfying the original condition, and with p, q, x all smaller than the original sides $(a^2-b^2),(2ab),(a^2+b^2)$—contradicting our assumption! Hence the assumption is false, and the proof is complete.

After this discussion, we've finally proved the problem we set out with. The method used here was first proposed by Fermat, and is known as "Fermat's method of infinite descent." The idea is simple: among the natural-number solutions of an equation there must be one attaining some minimum (e.g., the smallest value itself, or the smallest sum or product), unless no solution exists at all (since the smallest natural number is 0). Now let's see how this connects to Fermat's Last Theorem. For n = 4, Fermat's Last Theorem states that $a^4+b^4=c^4$ has no positive integer solutions. By Lemma 3, we have $a^2=x^2-y^2,b^2=2xy,c^2=x^2+y^2$; looking at the third equation, we've already proven above that there cannot exist two squares whose sum and difference are both squares—so a special case of Fermat's Last Theorem is proved! More generally, Fermat's Last Theorem fails to have solutions for every multiple of 4!

Now let's revisit the title of this post. In its most general form, the question is:

Does there exist a right triangle with area n whose three side lengths are all rational, where n is a positive integer?

This looks simple, but it is actually a very hard problem. To this day it can only be said to be "essentially" solved, not "completely" solved—in this sense its difficulty may even rival that of Fermat's Last Theorem. For certain specific values of n, we can prove that solutions exist or don't exist. Above, we proved the case n = 1; in fact, for n = 2, 3, 4 there are no solutions, while for n = 5, 6, 7 there are solutions. The following webpage lists the values of n for which the problem above has a solution:

https://oeis.org/A003273

Such n are called congruent numbers.

Using the discussion method above, it's not hard to see that when n is a prime number, this problem is equivalent to the following equation having positive integer solutions:

$$\begin{aligned}n\cdot x^2+y^2=z^2 \\ |n\cdot x^2-y^2|=w^2\end{aligned}$$

So how was this problem "essentially solved"? The answer is Tunnell's theorem. It states that when n is odd, if the number of integer solutions to the equation $n=2x^2+y^2+8z^2$ is twice the number of integer solutions to the equation $n=2x^2+y^2+32z^2$, then n is a congruent number; when n is even, if the number of integer solutions to the equation $n=2x^2+y^2+8z^2$ is twice the number of integer solutions to the equation $n=2x^2+y^2+32z^2$, then n is a congruent number; and if the number of integer solutions to the equation $n=8x^2+2y^2+16z^2$ is twice the number of integer solutions to the equation $n=8x^2+2y^2+64z^2$, then n is also a congruent number.

A problem that remained unsolved for many years has, to date, only received a "decision procedure"—we still cannot give a "constructive proof" (one that would directly produce the side lengths for a given n). This shows just how hard the problem is. Let's hope that someday a mathematician will crack this problem completely, once and for all! In any case, a hard problem is a hen that lays golden eggs!