, 2011

We will show the claim by assuming the contrary.
Assume that there exist non-constant polynomials $G(x, y)$ and $H(x, y)$, with real coefficients, such that
$$P(x, y) = G(x, y) \cdot H(x, y), \qquad (1)$$
where $P(x, y) = x^n + x y + y^n$, $n \in \mathbb{N}^*$.
Present $G(x, y)$ and $H(x, y)$ in the form of polynomials in $x$:
$$\begin{aligned} G(x, y) &= g_m(y) \cdot x^m + g_{m-1}(y) \cdot x^{m-1} + \dots + g_1(y) \cdot x + g_0(y), \quad m \in \mathbb{N}; \\ H(x, y) &= h_k(y) \cdot x^k + h_{k-1}(y) \cdot x^{k-1} + \dots + h_1(y) \cdot x + h_0(y), \quad k \in \mathbb{N}; \end{aligned}$$
where $g_i(y)$, $i=0,\ldots,m$, and $h_j(y)$, $j=0,\ldots,k$, are real polynomials in $y$.
It follows from (1):
$$m + k = n, \qquad (2)$$
$$\text{For } n \ge 2, \ g_m(y), h_k(y) \text{ are constant polynomials and hence are not divisible by } y. \quad (3)$$
$$\text{Since } G(x, y) \text{ and } H(x, y) \text{ are non-constant, by means of (3), if } n \ge 2, \text{ then } m, k \ge 1. \quad (4)$$
* If $n = 1$. Then, according to (2), we have $m + k = 1$. Consequently, $m = 0$ and $k = 1$, or $m = 1$ and $k = 0$.

Assume that $m = 0$ and $k = 1$. (The case $m = 1$ and $k = 0$ is treated similarly). Then we have
$$(y+1)x + y = g_0(y) h_1(y)x + g_0(y) h_0(y).$$
Consequently $g_0(y)(h_1(y) - h_0(y)) = 1$. Thus, $g_0(y)$ is a constant polynomial, contradicting the assumption that $G(x, y)$ is non-constant.

* If $n \ge 2$.
Let $i_0$ and $j_0$ be the least indices such that $g_{i_0}(y)$ and $h_{j_0}(y)$ are polynomials not divisible by $y$.
Clearly, the coefficients of $x^{i_0+j_0}$ in the expansion of $G(x, y) H(x, y)$ are
$$g_0(y) h_{i_0+j_0}(y) + g_1(y) h_{i_0+j_0-1}(y) + \dots + g_{i_0}(y) h_{j_0}(y) + g_{i_0+1}(y) h_{j_0-1}(y) + \dots + g_{i_0+j_0}(y) h_0(y)$$
It follows from the definition of $i_0$ and $j_0$ that the above coefficients are not divisible by $y$. Thus, by (1), with the remark that the coefficient of $x^n$ in $P$ is the unique one not divisible by $y$, we conclude $i_0 + j_0 = n$. Hence $i_0 = m$ and $j_0 = k$. Together with (4) we have either $m = 1$ or $k = 1$, as if otherwise, $m, k > 1$, by comparing the coefficients of $x$ on both sides of (1) we would have $y = g_0(y) h_1(y) + g_1(y) h_0(y) \neq y^2$, a contradiction.

Assume $m = 1$. (The case $k = 1$ is treated similarly). Then we have
$$x^n + x y + y^n = (a x + g_0(y))(b x^{n-1} + h_{n-2}(y) x^{n-2} + \dots + h_1(y) x + h_0(y)), \quad (5)$$
where $a, b$ are real constants with $b = 1$.
By (5) we have $y^n = g_0(y) h_0(y)$. Consequently $g_0(y) = a' y^s$, where $s \in \mathbb{N}^*, s \le n$ and $a'$ is a real constant, different from $0$.
Put $c = -\frac{a'}{a}$, we have $c \ne 0$. Plug $x = c y^s$ in (5), we obtain
$$c^n y^{s n} + c y^{s+1} + y^n = 0. \quad (6)$$
+ If $s = 1$ and $n = 2$, we obtain from (6): $(c^2 + c + 1)y^2 = 0$. Consequently $c^2 + c + 1 = 0$, a contradiction.
+ If $s = 1$ and $n > 2$, we obtain from (6): $(c^n + 1)y^n + c y^2 = 0$, a contradiction (since $c \ne 0$).
+ If $s \ge 2$ and $n \ge 2$ then $s n > n$ and $s n > s + 1$. Hence (6) is contradictory, since $c \ne 0$.

* Thus, in conclusion, the assumption at the very beginning is wrong and we thereby verify the claim of the problem.