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 sid is the square root of the sum $\,a = \sqrt{b+c}$.
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.
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 $b=a+1$, $b^2 - a^2 = a + b$