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).

Now, we found another way to generate Pythagorean Triples.

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4 thoughts on “Pythagorean Triple 2: Generating Pythagorean Triples

  1. This is super neat! I covered Pythagoren Triples in geometry this year and there wasn’t much I did with them; I just explained what they are and how they can be recognized. My students didn’t seem to dislike Pythagorean Triples but they didn’t seem to understand or feel a purpose for them either. I think that it would be neat to explore this with students and see if they could notice this pattern and generate a way to find Pythagorean Triples!

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