Fermat proved that if $(a, b, c)$ is a Pythagorean triple, then $(b, c, d)$ cannot be a Pythagorean triple.
Suppose $(a, nb, c)$ form a Pythagorean triple. Can $(b, c, d)$ be a Pythagorean triple? For which $n$ is this possible?
Another formulation is the following Diophatine system of equations:
$$a^2 + n^2b^2 = c^2$$ $$c^2 + b^2 = d^2$$
Fermat's theorem is then that $n$ cannot equal 1.
Searching with a computer for $d < 50,000$ and $n < 100$, I have found solutions for $n = 6, 16, 30, 48, 84$. For example, $n = 6$ has the solution $(a, b, c, d) = (99, 28, 195, 197)$ because $(99, 6*28, 195)$ and $(28, 195, 197)$ are both triples.
I am especially interested if there are any $n \neq 1$ that can be ruled out.