<s>[INST] Short proof for the determinant of a $4$ by $4$ matrix 
Prove that $\det \begin{bmatrix}x&y&z&t\\-y&x&-t&z\\-z&t&x&-y\\-t&-z&y&x\end{bmatrix} = (x^2+y^2+z^2+t^2)^2$

I'm looking for an elegant proof that doesn't involve bruteforce.
Since the answer is given, I'm thinking we can argue that the determinant here is a homogeneous polynomial $P(x,y,z,t)$ with degree $4$, that is invariant under $x\to -x$ and permutations of $x,y,z,t$.
As a result, $P(x,y,z,t) = \lambda (x^4+y^4+z^4+t^4) + \delta (x^2y^2+x^2z^2 + x^2t^2+y^2z^2 + y^2t^2 + z^2t^2)$
$\lambda$ and $\delta$ can be found by computing $P(0,0,0,1)$ or some such.

The problem is, it doesn't look easy to prove that $P$ doesn't change under permutation of $x,y,z,t$, neither that it's invariant when the variables are negated.

Can you suggest another short proof, or prove the two claims above ?
 [/INST] Hint:
$$
\begin{pmatrix}
   x &  y &  z &  t \\
  -y &  x & -t &  z \\
  -z &  t &  x & -y \\
  -t & -z &  y &  x
\end{pmatrix}
\begin{pmatrix}
   x & -y & -z & -t \\
   y &  x &  t & -z \\
   z & -t &  x &  y \\
   t &  z & -y &  x
\end{pmatrix} = (x^2+y^2+z^2+t^2)
\begin{pmatrix}
   1 & 0 & 0 & 0 \\
   0 & 1 & 0 & 0 \\
   0 & 0 & 1 & 0 \\
   0 & 0 & 0 & 1
\end{pmatrix} 
$$
</s>