Go I Know Not Whither and Fetch I Know Not What Next day: apparently my original question is harder, by far, than the other bits. So: it is a finite check, I was able to confirm by computer that, if the polynomial below satisfies $$ f(a,b,c,d) \equiv 0 \pmod {27}, \;\; \mbox{THEN} \; \; a,b,c,d \equiv 0 \pmod 3,  $$ and if
$$ f(a,b,c,d) \equiv 0 \pmod {125}, \;\; \mbox{THEN} \; \; a,b,c,d \equiv 0 \pmod 5,  $$ 
ORIGINAL:
$f$ is a polynomial in four variables. Take matrices
$$  
1 =
\left( 
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 
\end{array}
\right),  
$$
$$
i =
\left( 
\begin{array}{rrrr}
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 
\end{array}
\right),
$$
$$  
j =
\left( 
\begin{array}{rrrr}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 
\end{array}
\right),  
$$
$$
k =
\left( 
\begin{array}{rrrr}
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 
\end{array}
\right),
$$
Then take
$$ f(a,b,c,d) = \det (a \cdot 1  + b \sqrt 3 i + c \sqrt 5 j + d \sqrt{15} k),   $$
$$ =a^4-6 a^2 b^2+9 b^4-10 a^2 c^2-30 b^2 c^2+25 c^4+120 a b c d-30 a^2 d^2-90 b^2 d^2-150 c^2 d^2+225 d^4$$.  Note that everything is commutative; 
$$ i^2 = 1, j^2 = 1, k^2 = 1; \; ij=ji=k, ki=ik=j,jk=kj=i.  $$
It is also possible to re-write this with the square roots absorbed into the definitions of $i,j,k.$
So, questions include: does it make sense to anyone that, as I checked by brute force, that if
$$  f(a,b,c,d) \equiv 0 \pmod {81}  $$
then $a,b,c,d \equiv 0 \pmod 3?$ Same for $625$ and $5.$ Need to think about how to check $5$ completely.
Finally, is it true that this thing represents the same numbers as $x^2 - 15 y^2,$ and what is such a thing called anyway? It might be a field norm, I dunno.
Oh, from a closed question at https://math.stackexchange.com/questions/931769/integer-solution-to-diophantine-equations which I found interesting.
http://en.wikipedia.org/wiki/Go_I_Know_Not_Whither_and_Fetch_I_Know_Not_What
EDIT: It turns out we may use $27$ in place of $81.$ Evidently explaining this is the hard part. Confirmed, anyway. See what I can do with $125$ instead of $625.$
EDIT 2: Figured out how to program it; if the polynomial is divisible by $125,$ each variable is indeed divisible by $5.$
 A: Yes, this is a field norm; it is the norm of $a + b \sqrt{3} + c \sqrt{5} + d \sqrt{15}$, from $K = \mathbb{Q}(\sqrt{3}, \sqrt{5})$ down to $\mathbb{Q}$. Note that $a+b \sqrt{3} + c \sqrt{5} + d \sqrt{15}$ acts on the basis $(1, \sqrt{3}, \sqrt{5}, \sqrt{15})$ by
$$a \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} + 
b \begin{pmatrix} 0 & 3 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 5 \\ 0 & 0 & 1 & 0 \end{pmatrix} + 
c \begin{pmatrix} 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 5 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} + 
d \begin{pmatrix} 0 & 0 & 0 & 15 \\ 0 & 0 & 5 & 0 \\ 0 & 3 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}.$$
Now conjugate by the matrix whose diagonal entries are $(1, \sqrt{3}, \sqrt{5}, \sqrt{15})$ to get your matrix. The entries are no longer rational, so I can't think of the result as describing the action on $K$, but the determinant is the same.
$\mathbb{Q}(\sqrt{15})$ has class number $2$ and $K$ is the class field. So, for a prime $p$ other than $2$, $3$, $5$, we have that $\pm p$ is a value of $x^2-15 y^2$ if and only if $p$ splits principally in $\mathbb{Q}(\sqrt{15})$ if and only if $p$ splits in $K$ if and only if $\pm p$ is a value of $f$. Also, neither $x^2-15 y^2$ nor $f$ can be $3 \bmod 4$, so the sign is the same in the two cases.
However, they don't take the same set of composite values. Look at $-119 = 7 \times 17$. We have $61^2 - 15 \cdot 16^2 = -119$, but, if $7 | f(a,b,c,d)$ then $7^2 | f(a,b,c,d)$. 
I found this by hunting for two primes which are non-principally split in $\mathbb{Q}(\sqrt{15})$. In terms of quadratic forms, which I know you love, I needed primes of the form $3 x^2 - 5 y^2$, and I found $7=3 \cdot 2^2 - 5$ and $-17 = 3 \cdot 6^2 - 5 \cdot 5^2$.  Then their product was of the form $x^2-15 y^2$. 
Since these primes split non principally in $\mathbb{Q}(\sqrt{15})$, they don't split further in the class field. (We can also directly compute $\left( \frac{3}{7} \right) = \left( \frac{3}{17} \right) = -1$.) So things divisible by one power of $7$ or $17$ are not norms from $K$.
