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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 $
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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.

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    $\begingroup$ 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. $\endgroup$
    – Joppy
    Dec 15 '20 at 16:14
  • $\begingroup$ Are there any other in order? $\endgroup$
    – Burt
    Dec 17 '20 at 1:18
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The identity $$(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2$$ shows you how Pythagorean triples arise.

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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.

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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 $3^2+4^2=5^2$.

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