A square number is a number $n$ with the property $S(n) :\equiv (\exists x)\ x^2 = n$.
A number $n \neq 0$ is a square number iff it has an odd number of divisors.
A hypotenuse number is a number $n$ with the property $H(n) :\equiv (\exists x, y)\ x^2 + y^2 = n^2$.
A number is a hypotenuse number iff it has a prime factor of the form $4k+1$.
A Pythagorean triple is a triple $(a,b,c)$ of numbers with the property $P(a,b,c) :\equiv a^2 + b^2 = c^2$.
A triple $(a,b,c)$ is Pythagorean iff
$$(*)\ \ \ (\exists x,y,z)\ \phi(x,y,z) \wedge a = x(y^2 - z^2) \wedge b = x(2yz) \wedge c = x(y^2 + z^2)$$
with $\phi(x,y,z) :\equiv x,y,z\geq 0, y > z, y, z$ co-prime and not both odd.
I wonder if the following property $R(n)$ has been defined before in its own right and what its official name is.
$$R(n) :\equiv (\exists x,y,z)\ P(x,y,z) \wedge x + y + z =n$$
A number $n$ is $R$ iff a regular $n$-gon with rigid sides can be transformed into a right triangle, possibly in different ways, thus being the perimeter of a right-angled triangle (thanks to Ronald Blaak for this formulation). ($R$ stands for "right-triangulizable".)
From $(*)$ it's obvious that $R(n)$ iff $(\exists x,y,z)\ \phi(x,y,z) \wedge 2xy(y + z) = n$.
Is there another characterization of $R$ involving prime factors (instead of arbitrary divisors of which two have to be co-prime)?
The final characterisation of $R(n)$ :
$R(n)$ iff $(\exists x,y,z) \phi(x,y,z) \land 2 x y (y+z)=n $
is correct but can be simplified. Using that $x,y,z > 0$ (excluding zero-length sides), the resulting values of $a,b,c$ should be positive in order result in a proper length and perimeter. This is only relevant for $a>0$ because it gives $y>z$.
Since the perimeter of these triangles is always even, we find $\frac{n}{2}=x y (y+z)$ has to be an integer, with $1 \leq z < y < y+z$, and it follows that $2 \leq y < y+z < 2 y$. In other words, the half-perimeter needs to have to at least two different factors $p,q$ such that $p < q < 2 p$.
This is also sufficient, since if $\frac{n}{2}=p q r$ with $r\geq 1$, we can take $x=r$,$y=p$, and $z=q-p$ to find an example of a right-angled triangle with perimeter $n$.
This criterion is simpler(more efficient), in the sense that it does not assume any triplets $(x,y,z)$ in advance, but is acting directly on the perimeter $n$ and also constructs a valid $(a,b,c)$ triplet from the input $n$ and its factorisation when possible.
I am not aware of a special name for these numbers, but the sequence can be found in the encyclopaedia of integer sequences A010814.