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!
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 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 $p=2m(m+n)$.
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