generating pythagorean triples Archives

Pythagorean Triple 2: Generating Pythagorean Triples

January 26, 2011 GB High School Mathematics, High School Number Theory

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 $(11, 60, 61)$ .

Now, we found another way to generate Pythagorean Triples.

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