A pythagorean triple consists of three positive integers such that $c^2=a^2+b^2$. The three integers 3, 4 and 5  is a well known pythagorean triple because $ 5^2=3^2+4^2 $.

A pythagorean triple is easy to work with when training math because you don´t have to use rational numbers, only integers, and therefore loved by math teachers 🙂

A list of pythagorean triples

(3, 4, 5)
(5, 12, 13)
(8, 15, 17)
(7, 24, 25)
(20, 21, 29)
(12, 35, 37)
(9, 40, 41)
(28, 45, 53)

This is only a small list since it exists an infinite amount of pythagorean triples.

Create your own pythagorean triples

You can actually fairly simple create your own triples by scaling up one set. You do this by multiplying each value of the triple with a positive integer.

For example the pythagorean triple (3, 4, 5) can be multiplied with 3:

$(3·3, 4·3, 5·3) = (9, 12, 15)$

Let´s check if  the pythagorean theorem still holds:

$ 9^2+12^2= 225$

$ 15^2=225 $

The both sides equals each other!

Proof that we can multiply a pythagorean triple with a constant and get a new triple

Let´s check if it holds when we use algebra.

Let´s assume that we have found a triple $(a,b,c)$ where $a^2 + b^2 = c^2 $

Multiply the triple $(a,b,c)$ with a constant k so that we get $ (k·a, k·b, k·c) $

We now want to prove that the following holds

$(k·a)^2+(k·b)^2  = (k·c)^2 $

Expand both sides

$k^2a^2 +k^2b^2 = k^2c^2 $

$k^2(a^2 + b^2) =k^2c^2 $

If we divide both sides with $k^2$ we get

$a^2 +b^2 = c^2$ which we assumed holds from the beginning.

So the left side equals the right side and the proof is done.