# $3^2 + 4^2 = 5^2$. Are there more of these equations?

I found this equation and it actually makes sense once you figure it out: $$3^2 + 4^2 = 5^2$$ $$9 + 16 = 25$$ $$25 = 25$$

Are there any more of these kinds of equations?

## Here's a list I made of all I can find:

• $$3^2 + 4^2 = 5^2$$
• $$12^2 + 5^2 = 13^2$$

These numbers are called Pythagorean triples, as they satisfy Pythagoras Theorem. More generally, they are solutions of the Diophantine equation $$a^2 + b^2 = c^2$$. Here are a few more:

• $$8^2 + 15^2 = 17^2$$
• $$24^2 + 7^2 = 25^2$$
• $$40^2 + 9^2 = 41^2$$

A larger list can be found on Wikipedia. There are a lot of formulae to generate Pythagorean triples. According to Euclid's formula, the numbers $$a = m^2 - n^2, b = 2nm, c = m^2 + n^2$$ form a pythagorean triple $$a^2 + b^2 = c^2$$ for all combinations of $$m$$ and $$n$$.

Note that multiples of $$a, b, c$$ i.e. $$ka, kb, kc$$ also satisfy the equation. When $$a, b, c$$ have a highest common factor of 1, the triple is called a primitive pythagorean triple.

You can find out more about pythagorean triples in the mentioned links.

• I can’t help but share one of the most interesting things I learned recently: all the primitive triples fit into a ternary tree rooted at (3,4,5), with fairly simple rules producing each subtree. Dec 15 '20 at 16:14
• Are there any other in order?
– Burt
Dec 17 '20 at 1:18

The identity $$(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2$$ shows you how Pythagorean triples arise.

There is an infinite amount of such numbers, for example for any integer $$n$$ we have that $$(3n)^2+(4n)^2=(5n)^2$$, but there is also an infinite amount of elements $$(x,y,z)\in\mathbb Z^3$$ such that $$\text{gcd}(x,y,z)=1$$ and $$x^2+y^2=z^2$$. You can find more on Wikipedia.

Although this isn't quite what you asked, it's interesting that $$3^2+4^2=5^2$$ is the only 'consecutive' Pythagorean triple. In other words, there are no other positive integer solutions to $$(n-1)^2+n^2=(n+1)^2 \, .$$ This can be proven in the following way: \begin{align} (n^2-2n+1)+n^2&=(n^2+2n+1) \\ 2n^2-2n+1&=n^2+2n+1 \\ n^2-2n+1&=2n+1 \\ n^2-4n+1&=1 \\ n^2-4n&=0 \\ n(n-4)&=0 \\ n&=0 \text{ or }n=4 \, . \end{align} While $$(-1)^2+0^2=1^2$$ does satisfy the equation, it is usually ignored because we are trying to find triples that can form the lengths of the sides of a triangle. Thus, we are left with $$3^2+4^2=5^2$$.