I have been trying to find a formula to find the number of right angled triangles where all the sides are integers and the perimeter is given. I have seen the formula x3/((1-x2)(1-x3)(1-x4)), but that gives me the amount of possible triangles with that perimeter, not the amount of right – angled triangles. Does anyone know any formulas to find the amount of right angled, integer triangles of a given perimeter? Thanks in advance!
[Math] How to calculate the total number of possible right angle triangles where the perimeter is given, and all sides are integers
number theorypythagorean triplestriangles
Related Solutions
We also know that $a=k(u^2-v^2)$, $b=2kuv$, $c=k(u^2+v^2)$ for some integers $k,u,v$, so $$ 2k^2(u^2-v^2)uv=2009k\cdot(u^2-v^2+2uv+u^2+v^2)$$ i.e. $$ k(u-v)v = 2009.$$ Now match the factors on the left with factors of $2009$.
From the Wikipedia article on Thales' theorem:
If you know the hypotenuse, you know the base of this diagram; if you also know the area, then you can compute the height (from $A=\frac12 bh$). Then you can draw a horizontal line at the required height; where it intersects the circle is the vertex with the right angle.
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Best Answer
To ensure the sides are integers, we use the famous parametrization: $$a=m^2-n^2$$ $$b=2mn$$ $$c=m^2+n^2$$
Then we have: ($p$ is perimeter) $$p=a+b+c=2m^2+2mn=2m(m+n)$$
When $p$ is odd, no integer solutions for $m, n$ exist.
When $p=2ab$, the number of solutions is the number of $b$ such that $a<b<2a$, which is equivalent to the number of $k$ such that $k$ is a factor of $p$ and $\sqrt{p/2}<k<\sqrt{p}$.
For even $p \le 10$, there is no solution.
The first solution occurs at $p=12$, we have $m=2, n=1$, and $(a,b,c)=(3,4,5)$.
EDIT: I realized that a solution rarely exists. If you want to find some $p$ such that a solution exists, I suggest you substitute arbitrary values of $m,n$ into the formula $p=2m(m+n)$.