# Generating Pythagorean Triples From One Leg

The question is simple: Find the longest possible hypotenuse in a right triangle with integer sides where the shortest side has length T. What I am asking is if there are any means to approach this besides testing pythagorean triples by brute force.

Hint: If the given side is $T$, there will exist an $n$ such that $(n+1)^2-n^2=2n+1>T^2$.
To find Pythagorean triples with a given side B, we begin with the function: $$n=\frac{B}{2m}\text{ trying values of }m\text{ from }\biggl\lceil\sqrt{\frac{B}{2}}\space\space\biggr\rceil\text{ to }\frac{B}{2}\text{ for integer values of }n.$$
For example triples with side $$B=20$$. $$m=\biggl\lceil\sqrt{\frac{B}{2}}\space\space\biggr\rceil=4\text{ to }\frac{B}{2}=10$$ In the search we find $$f(5,2)=(21,20,29)\text{ and }f(10,1)=(99,20,101).$$
We can find different triples with the same odd legs, if they exist, using this function of $$(m,A)$$: $$\text{We can let }n=\sqrt{m^2-A}\text{ where m varies from }\lceil\sqrt{A}\space\rceil\text{ to }\frac{A+1}{2}$$ Sometimes there is only one match such as the smallest $$(3,4,5)$$ where $$m_{min}=m_{max}=2.$$ At other times, there are many matches such as for $$A=105$$ where $$m_{min}=11$$ and $$m_{max}=53.$$ and we find $$(m,n)=(11,4), (13,8), (19,16),\text{ and }(53,52).$$ For these respective values of $$(m,n)$$ we have:
$$A,B,C=105,38,137\quad A,D,E=105,208,233\quad A,F,G=105,608,617\quad A,H,I=105,5512,5513$$ These formulas find matching sides (if they exist) in a [small] finite search. When you find the largest opposite a given side A or B, that will be the one with the longest hypotenuse.