More on Pythagoras

Posted by Graham Wheeler on Sunday, January 17, 2010

Pythagoras is known for two great contributions to mathematics – he established the need for formal proofs instead of just conjecture and rules of thumb, and he established the existence of the irrationals. In popular culture of course, Pythagoras is more well known for the Pythagorean theorem – that the square of the hypotenuse of a right angled triangle is the sum of the squares of the other two sides – but this was one of the oldest known results in mathematics and in fact predates Pythagoras by as much as a thousand years. Nonetheless, Pythagoras provided a formal proof, and the result led him to ask whether there were rational numbers that worked in the case where the hypotenuse had the length two. That is, what was the fractional representation of the square root of two? Pythagoras managed to show there was none, leading to the discovery of the irrational numbers.

His proof was quite simple. Assume that \(\sqrt{2}\) can be expressed as a fraction \(\frac{a}{b}\) of two whole numbers \(a\) and \(b\). Assume these are the two smallest such whole numbers – that is, they have no common divisor allowing the fraction to be further reduced. Then:

$$\frac{a^2}{b^2} = 2$$


$$a^2 = 2 b^2$$

The right hand side is even, which means \(a^2\) is even, and thus \(a\) must be even, or \(a=2m\) for some \(m\). But then:

$${(2m)}^2 = 2b^2$$


$$2m^2 = b^2$$

So the left hand side is even, and thus \(b\) must be even. But if both \(a\) and \(b\) are even, then they have a common factor 2, which contradicts the assumption that they were the smallest such numbers.

There are, of course, infinitely many cases where the Pythagorean triangles have sides with rational length, and for that matter, integer length. In a recent post I mentioned the method described by Diophantus that inspired Fermat’s last theorem. It is easy to derive this method. Assume \(a\) and \(b\) are relatively prime, and that \(b\) is odd. Then:

$${(a+b)}^2 = a^2 + b^2 + 2ab > a^2 + b^2 = c^2$$

Thus \(c < a+b\). So we can write \(c = a+b-d\) for some positive \(d\). Then:

$$a^2 + b^2 = {(a+b-d)}^2 = a^2 + b^2 + d^2 + 2ab -2ad -2bd$$


$$d^2 = 2ad + 2bd - 2ab$$

From this we can see \(d\) must be even. Let \(d=2m\); then:

$$4m^2 = 4am+4bm-2ab$$



Thus \(ab\) is even, and as we assumed \(b\) is odd, \( a\) must be even. Since the sum of an even and an odd is an odd, \( c^2\) must be odd and so \(c\) must be odd.

So \(a\) is even, and \(b\) and \(c\) are odd. We can write \(b=s-t\) and \(c=s+t\) for some integers \(s\) and \(t\). Then:

$$a^2 + {(s-t)}^2 = {(s+t)}^2$$


$$a^2 + (s^2 - 2st + t^2) = (s^2 + 2st + t^2)$$


$$a^2 = 4st$$

So \(st\) must be a perfect square. We can write \(s=u^2\) and \(t=v^2\). Thus:

$$a^2 = 4u^2v^2$$



So we have shown that a Pythagorean triple takes the form:

$$( 2uv, u^2-v^2, u^2 + v^2 )$$