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I have 1 right triangle of dimensions $\sqrt75$$, 11, 14$. I'd like to know how to quickly obtain the other right triangles with $\sqrt75$ as a leg, and two integers as the hypotenuse and the other leg (as per the Pythagorean theorem). It is to my understanding that these triangles are all connected somehow geometrically and, consequently, algebraically. Are the necessary techniques for quickly obtaining them related to: https://en.wikipedia.org/wiki/Spiral_of_Theodorus and/or the proof using differential techniques shown here: https://en.wikipedia.org/wiki/Pythagorean_theorem ?

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  • $\begingroup$ I don't see how this may be related to Theodorus' spiral. Anyway, using Pithagoras' theorem you can prove (see my answer) that there exist exactly 3 such right triangles. $\endgroup$ – Crostul Mar 19 '16 at 23:44
  • $\begingroup$ Got it, thanks! I just thought there might be a way figure out the other sides without having to determine the divisors of the common side. $\endgroup$ – user3108815 Mar 19 '16 at 23:46
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Factor $75=3 \cdot 5^2$. This implies that $75$ has 6 divisors, namely $1,3,5,15,25,75$.

Your problem is equivalent on looking for all integer solution to $$x^2+75=y^2$$ Now, write this equation as $$(y+x)(y-x)=75$$ so that you have to solve 6 different linear systems $$\left\{ \begin{matrix} y &+&x& =& a \\ y&-&x&=& 75/a\end{matrix} \right.$$ where $a$ is a divisor of $75$. For example, for $a=25$ you get the solution $x=11, y=14$.

The other solutions are $(37,38)$ and $(5,10)$ (and other 3 negative solutions, which must be discarded).

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  • $\begingroup$ Ah but one has to find the divisors of a number first? $\endgroup$ – user3108815 Mar 19 '16 at 23:43
  • $\begingroup$ Since you want $x,y$ to be integers, it is necessary to consider only divisors of $75$ ($a=x+y$ must be an integer, and the same for $75/a=x-y$). If you are not interested in having sides with integer measure, then every possible value of $a$ is allowed, giving you infinitely many systems, hence infinitely many solutions. But these three are the unique ones with $x,y$ integers. $\endgroup$ – Crostul Mar 19 '16 at 23:46

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