Can the Area of a Rational Right Triangle be an Integer?

This is an ancient and interesting problem, but before introducing it, let us first look at a simple question:

Can the area of a right triangle with integer sides be a perfect square?

The answer is no. We can examine some examples to verify this. For instance, a right triangle with side lengths 3, 4, and 5 has an area of 6, and 6 is not a square number. Another example is a right triangle with side lengths 5, 12, and 13, which has an area of 30, and 30 is also not a square number... Of course, the ultimate goal of mathematics is to require a strict proof rather than simple examples; otherwise, it can only be called incomplete induction, and what is derived is a conjecture rather than a “theorem,” much like the famous “Goldbach’s Conjecture”... In this article, we will attempt to prove this proposition.

We will later discover that this problem is equivalent to the following:

Even more surprisingly, this problem is also equivalent to the statement that the equation x^4 + y^4 = z^4 has no integer solutions. In other words, we are going to prove Fermat’s Last Theorem for n=4!

Now, starting from the initial problem, let us attempt the proof. First, we must prove three lemmas:

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

2. If the product of n pairwise coprime integers is a perfect square, then each of these n integers is a perfect square.

3. Any Pythagorean triple can be expressed in the form (a^2-b^2)^2 + (2ab)^2 = (a^2+b^2)^2, where a and b are positive integers.

The proofs of these lemmas are not difficult and can be derived using some basic theorems of elementary number theory; therefore, they will not be proven in detail here. As a hint for the second lemma: if a square number has a prime factor p, then the count of p must be even. If we distribute these factors among n integers and p is assigned to two different integers, then those two integers must have a common divisor p, which contradicts the premise that the “n integers are pairwise coprime.” Thus, all factors of p must be assigned to a single integer. This applies to every prime factor of the square number, so these n integers are all products of even numbers of prime factors, meaning they are all square numbers.

Now, let us consider the right triangle with integer sides. We use proof by contradiction. Assume there exists a case where the area is a square number. Then there must exist a right triangle with coprime legs whose area is a square number, and this triangle has the shortest perimeter among all such triangles (in a set of integer-sided triangles, there always exists one with a minimum perimeter). We know that all Pythagorean triples can be represented as (a^2-b^2)^2 + (2ab)^2 = (a^2+b^2)^2. Since we have assumed the two legs are coprime, a and b must be coprime integers, one odd and one even, with a > b.

The area of the triangle is \frac{1}{2} \cdot 2ab \cdot (a^2-b^2) = ab(a-b)(a+b). By our assumption, this is a square number. From Lemma 1, a, b, (a+b), (a-b) are pairwise coprime. From Lemma 2, a, b, (a+b), (a-b) are all square numbers. Let a=x^2 and b=y^2. We denote: \begin{aligned} x^2 + y^2 &= z^2 \\ x^2 - y^2 &= w^2 \end{aligned} It is easy to see that y must be even.

Multiplying the two equations gives x^4 - y^4 = (zw)^2, which indicates that (zw), y^2, x^2 also form a Pythagorean triple. By Lemma 3, let zw = p^2 - q^2, y^2 = 2pq, and x^2 = p^2 + q^2, where p and q are positive integers. Then there exists a right triangle with side lengths p, q, x whose area is \frac{1}{2}pq = (\frac{y}{2})^2. This satisfies the initial condition, yet p, q, x are all smaller than the original sides (a^2-b^2), (2ab), (a^2+b^2), which contradicts our assumption! Thus, the assumption is false. Q.E.D.

After this discussion, we have finally proven the initial problem. 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 a set that satisfies a certain minimum value (such as the value itself, their sum, or their product); otherwise, the solutions cannot exist (since the smallest natural number is 0). Now, what does this have to do with Fermat’s Last Theorem? When n=4, Fermat’s Last Theorem states that a^4 + b^4 = c^4 has no positive integer solutions. According to Lemma 3, we have a^2 = x^2 - y^2, b^2 = 2xy, and c^2 = x^2 + y^2. Focusing on the third equation, we have already proven that there do not exist two square numbers whose sum and difference are both square numbers. Thus, a special case of Fermat’s Last Theorem is proven! More generally, Fermat’s Last Theorem does not hold for all n that are multiples of 4!

Now, let us look at the title of this article again. In fact, the most general formulation of this problem is:

This looks simple, but it is actually a major difficult problem. So far, this problem can only be considered “basically” solved rather than “completely” solved. In this sense, its difficulty can even be compared to Fermat’s Last Theorem. For certain specific values of n, we can prove whether a solution exists. We have already proven the case for n=1. In fact, there are no solutions for n=2, 3, 4, while solutions exist for n=5, 6, 7. The following website lists the values of n for which the problem has a solution:

In this context, n is called a congruent number.

Using the discussion method above, it is not difficult to find that when n is a prime number, this problem is equivalent to the existence of positive integer solutions for the following equations: \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 “basically 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; 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.

This problem, which remained unresolved for many years, has so far only received a “decisional proof” and cannot yet provide a “constructive proof” (i.e., one that can provide specific side lengths based on n). This highlights the difficulty of the problem. May future mathematicians completely conquer this issue! Regardless, a difficult problem is a hen that lays golden eggs!

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