How can $f(x,y)= x^4+x^3y+x^2y^2+xy^3+y^4$ be factorized into a product of two polynomials? Let $x,y$ be 2 coprime integers. I assume the following polynomial:$$f(x,y)= x^4+x^3y+x^2y^2+xy^3+y^4$$ is not irreducible. So there must be at least 2 other polynomials of degree $\leq 4$ such that: $$f(x,y)=g(x,y)h(x,y)$$
How can one find $g(x,y)$ and $h(x,y)$?
 A: The important thing to notice
is that
$f(x, y)$
is homogeneous;
every term
of the form
$x^ay^b$
has the same value of
$a+b$.
Therefore,
dividing by
$y^n$,
where $n$ is this sum,
$\begin{array}\\
f(x,y)
&= x^4+x^3y+x^2y^2+xy^3+y^4\\
&= y^4 \left((x/y)^4+(x/y)^3+(x/y)^2+(x/y)+1 \right)\\
&= y^4 \left(r^4+r^3+r^2+r+1 \right)
\qquad\text{where } r=x/y\\
\end{array}
$
Now the problem is reduced
to factoring
$r^4+r^3+r^2+r+1
$.
A: According to marty cohen, we have
\begin{align*}
r^4+r^3+r^2+r+1&=\left(r^2+\frac{1}{2}r+1\right)^2-\frac{5}{4}r^2\\
&=\left(r^2+\frac{1+\sqrt{5}}{2}r+1\right)\left(r^2+\frac{1-\sqrt{5}}{2}r+1\right)\\
\end{align*}
Then
$$x^4+x^3y+x^2y^2+xy^3+y^4=\left(x^2+\frac{1+\sqrt{5}}{2}xy+y^2\right)\left(x^2+\frac{1-\sqrt{5}}{2}xy+y^2\right)$$
A: One way of showing irreducibility over the rationals is to reduce the problem to a more familiar one. If $f(x,y)$ were reducible over the rationals, then the polynomial $g(x)=f(x,1)=x^4+x^3+x^2+x+1$ would be reducible over the rationals. But that is not the case. One way to show this is to use the Eisenstein Criterion on $g(x+1)$.  Another way is to note that if $g(x)$ split over the rationals, it would split over the integers. But $g(x)$ has no rational root, by the Rational Root Theorem. So we try to split $g(x)$ as a product of quadratics with integer coefficients. The only possibility is something like $(x^2+ax\pm 1)(x^2+bx\pm 1)$, and a little chasing shows no integers $a,b$ work.
Remark: One can split $g(x)$, and therefore $f(x,y)$ as a product of quadratic polynomials with real coefficients. This can be done by using the fact that $g(x)$ splits over the complex numbers as $(x-a_1)(x-a_2)(x-a_3)(x-a_4)$ where $a_k=e^{2\pi i k/5}$.
The polynomial $(x-a_1)(x-a_4)$ has real coefficients, as does $(x-a_2)(x-a_3)$. 
A: Here's an efficient route, in the spirit of the remark in André Nicolas' answer:
On one hand, the product of $f(x, y)$ with $(x - y)$ telescopes:
$$(x - y) (x^4 + x^3 y + x^2 y^2 + x y^3 + y^4) = x^5 - y^5.$$
On the other,
$$x^5 - y^5 = \prod_{k = 0}^4 (x - \zeta^k y),$$
where $\zeta := e^{2 \pi i / 5},$ and so over $\Bbb C$ (in fact, over $\Bbb Q[\zeta]$), $f$ factors as
$$f(x, y) = \prod_{k = 1}^4 (x - \zeta^k y).$$
(Note that the lower limit in this product is not the same as in the earlier one.)
The irreducible factors of $f$ over $\Bbb R$ are thus the (quadratic) products of the conjugate factors, namely,
\begin{align}(x - \zeta^2 y) (x - \bar{\zeta}^2 y) = (x - \zeta^2 y) (x - \zeta^3 y) = x^2 - 2 (\Re (\zeta^2)) xy + y^2 &= x^2 + \phi xy + y^2 \\ (x - \zeta y) (x - \bar{\zeta} y) = (x - \zeta y) (x - \zeta^4 y) = x^2 - 2 (\Re \zeta) xy + y^2 &= x^2 + \tilde{\phi} xy + y^2 . \end{align}
Here, $\phi := \frac{1}{2}(1 + \sqrt{5})$ is the Golden Ratio and $\tilde{\phi} := \frac{1}{2}(1 - \sqrt{5})$ is its conjugate (in $\Bbb Q[\sqrt{5}]$). In particular, since some of the irreducible coefficients of the factors of $f$ over $\Bbb R$ are irrational, $f$ is irreducible over $\Bbb Q$.
