Vector Geo Problem I am lost in. Question:
There exists positive integers $x, y, z$ where $\gcd(x,y,z)=1$, so that for any Pythagorean triple $(a,b,c)$
$$\begin{pmatrix} x & y & y \\ y & x & y \\ y & y & z \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix}$$
is also a Pythagorean triple.
So I tried to bash this question, finding that there should a way to solve it by multiplying the matrices and somehow factoring it.
So I got
$$\begin{pmatrix} xa + y(b+c) \\ y(a+c)+xb \\ y(a+b)+zc \end{pmatrix}$$
then I tried to put it through the Pythagorean theorem, but I could not see a good path to a solution.
Am I missing something?
Thanks in advance.
 A: The obvious thing to do would be to set $(a,b,c)$ to the smallest primitive Pythagorean triples $(3,4,5)$, $(5,12,13)$ and $(8,15,17)$ to get the system of equations
\begin{align}
(3x+9y)^2+(4x+8y)^2 & = (7y+5z)^2\\
(5x+25y)^2+(12x+18y)^2 & = (17y+13z)^2\\
(8x+32y)^2+(15x+25y)^2 & =(23y+17z)^2
\end{align}
and solve the system of equations. This can be done by computer.
However, if there is a solution, we expect to be based in a purely algebraic identity, which means that it will have to work even with degenarate Pythagorean triples. So we try setting $(a,b,c)$ equal to $(1,0,1)$ and $(1,0,-1)$, which give us the equations
\begin{align}
(x+y)^2+(2y)^2 & = (y+z)^2\\
(x-y)^2+0^2 & = (y-z)^2
\end{align}
This is a simpler system of equations that we can solve by hand to get the families of solutions:
$x=t$, $y=0$, $z=t$ with $t \in \mathbb R$
$x=t$, $y=0$, $z=-t$ with $t \in \mathbb R$
and
$x=t$, $y=2t$, $z=3t$ with $t \in \mathbb R$
(This solutions are actually the same of the first hard system of equations.)
Since we are interested in solutions with coprime positive integers, the only solution is $x=1$, $y=2$, $z=3$.
It's easy to verify that this is in fact a solution that work with any Pythagorean triple.
