Pythagorean Triple 2: Generating Pythagorean Triples

In the previous math article, we have shown that there are infinitely many Pythagorean triples. In this article, we are going to discuss a very short but effective strategy in generating Pythagorean Triples.

A Pythagorean triple is the integer triple $(a,b, c)$ satisfying the Pythagorean equation $a^2 + b^2 = c^2$ .

Observe the Pythagorean triples $(3, 4, 5)$ and $(5, 12, 13)$ . We can see that the hypotenuse is greater than the longer side by $1$ . From the pattern, we can form the Pythagorean triples $(a, b, b + 1)$ satisfying the equation $a^2 + b^2 = (b + 1)^2$ .

Right triangle with side length 3, 4 and 5 units.

Solving the equation we have $a^2 = 2b + 1$ , which implies that $a = \sqrt{2b +1}$ . Now, $2b + 1$ is always odd (can you see why?). It follows that in order for $a$ to be an integer, $2b + 1$ must be a perfect square. This means, that we are sure that $a$ is an integer, if $2b + 1$ is an odd perfect square.

From here, we can generate infinitely many examples of Pythagorean triples. For example, $49$ is an odd perfect square. So plugging it in the equation we have, $2b + 1 = 49$ , then, we have the triple $(7, 24, 25)$ , another Pythagorean Triple. If we let $2b + 1 = 121$ , then we have the triple $(11, 60, 61)$ .