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So I have search to the best of my abilities but cannot find mathematically why this is true and if it is called something specific, the closest thing would be Pythagorean Triples but this is not the case although a (3,4,5) is a triple. I will do my best to explain and apologize in advance as math is not my strength.

So in any right triangle I could find the following holds true. If (a,b,c) and b differs -1 from c the other angle is the sum of a = sqrt{b+c}.

For example:

(3,4,5) a = sqrt{4+5} = sqrt{9} = 3 = sqrt{9} = sqrt{25-16} = sqrt{5^2 - 4^2}

(a,17,18) a = sqrt{17+18} = sqrt{35} = sqrt{324-289} = sqrt{18^2 - 17^2}

The only way I understand it is the difference between two squared numbers who base differs by 1 will always be the sum of both numbers.

So assuming b is always a+1, b^2 - a^2 = a + b. Any explanation or pointing to where I can read about more would be much appreciated.

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Yes, that is true. Given $b = (a+1)$

$$ b^2-a^2\\ =(a+1)^2 - a^2\\ = a^2+2a +1 - a^2\\ = 2a + 1 \\= a + (a + 1)\\= a+b $$

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For all $a,b$ $$b^2 - a^2 = (b+a)(b-a)$$ Assuming as you do that $b=a+1$ $$b^2 - a^2 = (b+a)(b-a) = (2a+1)(1)=2a+1$$ and $$a+b=a+a+1=2a+1$$ so yes, when $b=a+1$, $b^2 - a^2 = a + b$

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There are relatively few triples where $A$ and $B$ differ by only $1$. In triples with sides under 212 quadrillion, there are only $19$ of them and one method of finding them is here.

Here are the first few shown as f(n,k) and they agree with the $proof$ in notovny's answer: $$3,4,5\Rightarrow4^2-3^2=16-9=7=3+4$$ $$21,20,29\Rightarrow21^2-20^2=441-400=41=21+20$$ $$119,120,169\Rightarrow120^2-119^2=14400-14161=239=120+119$$ If you want $C-B=1$, these functions will generate them all for an $k\in\mathbb{N}$

$$A=2k+1\qquad B=2k^2+2k\qquad C=2k^2+2k+1$$

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