For which $n$ can $(a, nb, c)$ and $(b, c, d)$ be Pythagorean triples? 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.

An Attempt
You have $a^2+\left(n^2+1\right)b^2=d^2$.  All integral solutions $(a,b,d)$ to this equation satisfies (1) $|a|=|d|$ and $b=0$, or (2) $d\neq 0$ and $\left(\frac{a}{d},\frac{b}{d}\right)=\left(\frac{\left(n^2+1\right)r^2-1}{\left(n^2+1\right)r^2+1},\frac{2r}{\left(n^2+1\right)r^2+1}\right)$ for some $r\in\mathbb{Q}$.
Now, if $b^2+c^2=d^2$, then (1) $|c|=|d|$ and $b=0$, or (2) $d\neq 0$ and $\left(\frac{b}{d},\frac{c}{d}\right)=\left(\frac{2s}{s^2+1},\frac{s^2-1}{s^2+1}\right)$ for some $s\in\mathbb{Q}$.  Thus, we have to find $r,s\in\mathbb{Q}$ such that $$\frac{r}{\left(n^2+1\right)r^2+1}=\frac{s}{s^2+1}\,.$$
Consequently, $$rs^2-\big(\left(n^2+1\right)r^2+1\big)s+r=0\,.$$
This means
$$\big(\left(n^2+1\right)r^2+1\big)^2-4r^2=q^2$$
for some $q\in\mathbb{Q}$.  Hence,
$$\left(n^2+1\right)^2r^4+2\left(n^2-1\right)r^2+1=q^2\,.$$
Therefore, we have
$$\left(\left(n^2+1\right)r^2+\frac{n^2-1}{n^2+1}\right)^2+\left(\frac{2n}{n^2+1}\right)^2=q^2\,.$$
We have again run into a Pythagorean problem.  Using a similar argument, the problem boils down to finding $(r,u)\in\mathbb{Q}\times\mathbb{Q}$ satisfying $u\neq 0$ and $$\frac{\left(n^2+1\right)^2r^2+\left(n^2-1\right)}{n}=\frac{u^2-1}{u}\,.$$
For example, $(n,r,u)=\left(6,\frac27,-\frac2{49}\right)$ produces $(n,a,b,c,d)=(6,99,28,195,197)$. I do not have any idea how to proceed further than this.  (To be honest, I haven't reduced the number of variables at all.  Being rationals, $r$ and $u$ each account for $2$ integer variables.)
Will Jagy's infinite family is produced by $$(n,r,s,q,u)=\left(2\left(t^2-1\right),\frac{t}{2t^2-1},t\left(2t^2-1\right),\frac{4t^6-4t^4+t^2-1}{\left(2t^2-1\right)^2},-\frac{2}{\left(2t^2-1\right)^2}\right)$$
for $t=2,3,\ldots$.
