Fermat's Only Published Proof

"Bachet: Find a triangle whose area is a given number." The area of a triangle in numbers cannot be a square. I am going to give a proof of this theorem, which I have discovered; and I did not find it without painful and laborious thinking about it, but this kind of proof will lead to marvellous progress in the science of numbers.

If the area of a triangle was a square there would be two fourth powers whose difference was a square; it would follow equally that there would be two squares whose sum and difference would be squares. Consequently there would be a square number, the sum of a square and the double of a square, with the property that the sum of the two squares that make it up is likewise a square. But if a square number is the sum of a square and the double of a square its root is likewise the sum of a square and the double of a square, which I can prove without difficulty. One concludes from this that this root is the sum of the two sides of a right angle in a triangle of which one of the square components forms the base and the double of the other square the height.

This triangle will therefore be formed of two square numbers whose sum difference are squares. But one will prove that the sum of these two squares is sma1ler than the first two of which one has likewise supposed that the sum and difference are squares. Therefore, if one has two squares whose sum and difference are squares one has at the same time, in integers, two squares enjoying the same property whose sum is less.

By the same reasoning, one has accordingly another sum smaller than that derived from the first, and continuing indefinitely one will always find smaller and smaller integers satisfying the same condition. But this is impossible because an integer being given there cannot be an infinity of integers which are smaller.

The margin is too narrow to receive the complete proof with all its developments.

In the same way I have discovered and proved that there cannot be a triangular number, except for one, which is also a fourth power.

Fortunately, just for once, he had found room for this mystery ib the margin of the very last proposition of Diophantus; this is how it goes.

Take a pythagorean triangle whose sides may be assumed mutually prime; then they can be written as \((2pq, p^{2} - q^{2}, p^{2} + q^{2})\) where \(p,\space q\) are mutually prime, \(p\gt q\), \(p - q\) is odd. Its area is \(pq(p + q)(p - q)\), where each factor is prime to the other three; if this is a square, all the factors must be squares. Write \(p = x^{2}\), \(q = y^{2}\), \(p + q = u^{2}\), \(p - q = v^{2}\), where \(u\), \(v\) must be odd and mutually prime. Then \(x\), \(y\) and \(z = uv\) are a solution of \(x^{4} - y^{4} = z^{2}\); incidentally, \(v^{2}\), \(x^{2}\), \(u^{2}\) are then three squares in an aritithmetic progression whose difference is \(y^{2}\). We have \(u^{2} = v^{2} + 2y^{2}\); writing this as \(2y^{2} = (u + v)(u - v)\), and observing that the g.c.d. of \(u + v\) and \(u - v\) is \(2\), we see that one of them must be of the form \(2r^{2}\) and the other of the form \(4s^{2}\), so that we can write \(u = r^{2} + 2s^{2}\), \(\pm v = r^{2} - 2s^{2}\), \(y = 2rs\), and consequently

\(x^{2} = \frac{1}{2}(u^{2} + v^{2}) = r^{4} + 4s^{4}.\)

Thus \(r^{2}\), \(2s^{2}\) and \(x\) are the sides of a pythagorean triangle whose area is \((rs)^{2}\) and whose hypotenuse is smaller than the hypotenuse \(x^{4} + y^{4}\) of the original triangle. This completes the proof "by descent".