Any positive integer?

We will demonstrate that for any $a,$ there exists a Pythagorean triple, i.e. a triple of positive integers $(x, y, z)$ such that $x^2 + y^2 = z^2,$ with $a$ as one of its values. We divide the problem into cases based on the parity of $a.$

If $a$ is even, then we can write $a = 2k,$ where $k$ is a positive integer greater than 1. Then, $(a, b, c) = (2k, k^2 - 1, k^2 + 1)$ is a Pythagorean triple, because $c^2 - b^2 = (c-b)(c+b) = 2(2k^2) = (2k)^2 = a^2.$

If $a$ is odd, then we can write $a = 2l + 1,$ where $l$ is a positive integer. Then, $(a, b, c) = (2l + 1, 2l^2 + 2l, 2l^2 + 2l + 1)$ is a Pythagorean triple, because $c^2 - b^2 = (c-b)(c+b) = 4l^2 + 4l + 1 = (2l + 1)^2 = a^2.$

Therefore, we can say that yes , any positive integer greater than 2 can be the side length of a right triangle with integer side lengths.