Primes $p=1\pmod 5$ Let $p=1\pmod 5$ be a prime. Then 
$p=x^4 + x^3z + x^2z^2 + 5xy^3 - 5xy^2z + xz^3 + 5y^4 - 10y^3z + 10y^2z^2 - 5yz^3 + z^4$
for integers $x, y,$ and $z$.
Is there a proof for this? 
For instance the fifth cyclotomic field $\mathbb Q (e^{i\frac{\pi}{5}})$ is of degree $4$ and class number one, so each prime $p=1\pmod 5$ is written as the norm of an element 
$aw^3+bw^2+cw+d$
(13:58) gp > norm(Mod(a*x^3+b*x^2+c*x+d,polcyclo(5)))
%41 = a^4 + (-b + (-c - d))*a^3 + (b^2 + (2*c + 2*d)*b + (c^2 - 3*d*c + d^2))*a^2 + (-b^3 + (-3*c + 2*d)*b^2 + (2*c^2 - d*c - 3*d^2)*b + (-c^3 + 2*d*c^2 + 2*d^2*c - d^3))*a + (b^4 + (-c - d)*b^3 + (c^2 + 2*d*c + d^2)*b^2 + (-c^3 - 3*d*c^2 + 2*d^2*c - d^3)*b + (c^4 - d*c^3 + d^2*c^2 - d^3*c + d^4))

where 
$w=e^{i\frac{\pi}{5}}$, a fifth root of unity. However, the element contains four parameters, whereas the form of interest contains three? Is it still suitable to represent all primes $p$ congruent to $1\pmod 5$?
 A: Take $w$ being a fifth root of unity and $K = \mathbb{Q}(w)$.  The ring of algebraic integers is $O := \mathbb{Z}[w]$ and ideal-class group is trivial (you can compute this for instance with Sage or Magma). That means that every ideal is in $O$ is principal.
If $p$ is 1 mod 5, then $\mathbb{F}_p$ contains a 5'th root of unity, and hence $x^5-1$ splits over $\mathbb{F}_p$. Then the minimal polynomial of $w$, which is $(x^5-1)/(x-1)$, also splits over $\mathbb{F}_p$. That implies that the ideal $(p) \subset O$ splits, it factors as a product of 4 ideals, each of which has Norm $p$.  Now take one of those 4 ideals. Since the class group is trivial, it follows that that ideal is principal, say, equal to $(g)$.
The Norm of $g$ is $\pm$ the Norm of the ideal $(g)$, so it is $\pm p$, but we can easily rule out $-p$ because no element of $K$ has negative Norm. So the Norm of $g$ is $p$.
Now the Norm of $h := x + (y-z) w - y w^2$ is precisely the polynomial you gave.   Notice that the $w^3$ term does not appear.
So the question then becomes: Given $g$, does there exist a unit $u \in O$ such that $ug$ is $w^3$-less? The group of units is isomorphic to $\mathbb{Z}/10 \oplus \mathbb{Z}$.  I suspect that in general the answer is no, and that this can be settled with a finite computation. Say the units are generated by $-1,w,u$. So now we have to check if there exist $k,n$ for which $\pm w^k u^n$ is $w^3$-less. The $\pm$ does not matter. Fix $k \in \{0,1,2,3,4\}$ and denote the $w^3$-coefficient of $w^k u^n g$ as $a_n$.  Then $a_n$ satisfies a linear recurrence relation, with constant coefficients, of order 4. The characteristic polynomial of this recurrence relation is the minpoly of $u$.
We have to show that $a_n \neq 0$ for any integer $n$.  One plan would be to find some prime number $q$ for which $a_n \not\equiv 0$ mod $q$ for all $n$ (the latter is a finite computation because the sequence $(a_n)$ repeats mod $q$.
Remains to do:  find suitable $p$ (some were mentioned in a comment). Then compute $a_n$ and find suitable $q$ (for each $k=0,\ldots,4$).
Added later:  Now that I actually did the computation, I find that it is not necessary to look for such $q$.  The group of units is generated by: $-1$ (order 2), $w$ (order 5), and $u$ = golden ratio $= -(w^2+w^3)$ (infinity order). This means that the recurrence relation is: $a_{n+2} = a_{n+1} + a_0$ but for those sequences it is very easy to verify if $0$ is a member of the sequence or not because for $n>>0$ the sequence grows as constant times $u^n$ while for $n<<0$ it grows as constant times $(-u)^n$.
So the computation is now as follows: $p = 1181$ and $g := w^3+3w^2-4w-2$ has Norm $p$. So the factorization of the ideal $(p)$ is $(g) (g_2) (g_3) (g_4)$ where $g_i$ is obtained from $g$ by replacing $w$ with $w^i$  (i.e. the $g_i$ are obtained from $g$ by applying elements of the Galois group).
Next, take $k=0$ and let $a_n$ be the $w^3$-coefficient of $w^k u^n g$. You only have to compute that for $n=0$ and $n=1$ because then all other $a_n$ for $n \in \mathbb{Z}$ are determined by the recurrence relation. One finds $a_0, a_1 = 1,-4$. Then the recurrence tells us: $\ldots,a_{-4},\ldots,a_4,\ldots = \ldots,17, -11, 6, -5, 1, -4, -3, -7, -10,\ldots$ and it is clear from $a_{n+2} = a_{n+1} + a_0$ that this sequence does not contain $0$.
Repeat this for $k=1,2,3,4$.  In principle we have to repeat this for $g_2$, $g_3$, and $g_4$ but this can be avoided with a short argument. In any case, this  reduces the problem to a finite computation.
