Vijetnam 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 $$P(x,y)=G(x,y)\cdot H(x,y),(1)$$ where $P(x,y)=x^n+xy+y^n$, $n\in {\mathbb{N}}^{\ast }$. Present $G(x,y)$ and $H(x,y)$ in the form of polynomials in $x$: $$\begin{array}{rl}G(x,y)& =g_m(y)\cdot x^m+g_{m-1}(y)\cdot x^{m-1}+\cdots +g_1(y)\cdot x+g_0(y),m\in \mathbb{N};\\ H(x,y)& =h_k(y)\cdot x^k+h_{k-1}(y)\cdot x^{k-1}+\cdots +h_1(y)\cdot x+h_0(y),k\in \mathbb{N};\end{array}$$ where $g_i(y)$, $i=0,\dots ,m$, and $h_j(y)$, $j=0,\dots ,k$, are real polynomials in $y$. It follows from (1): $$m+k=n,(2)$$ $$\text{For }n\ge 2,g_m(y),h_k(y)\text{ are constant polynomials and hence are not divisible by }y.(3)$$ $$\text{Since }G(x,y)\text{ and }H(x,y)\text{ are non-constant, by means of (3), if }n\ge 2,\text{ then }m,k\ge 1.(4)$$ 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 $$g_0(y)h_{i_0+j_0}(y)+g_1(y)h_{i_0+j_0-1}(y)+\cdots +g_{i_0}(y)h_{j_0}(y)+g_{i_0+1}(y)h_{j_0-1}(y)+\cdots +g_{i_0+j_0}(y)h_0(y)$$ 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 $y=g_0(y)h_1(y)+g_1(y)h_0(y)\ne y^2$, a contradiction.

Assume $m=1$. (The case $k=1$ is treated similarly). Then we have $$x^n+xy+y^n=(ax+g_0(y))(b{x}^{n-1}+h_{n-2}(y)x^{n-2}+\cdots +h_1(y)x+h_0(y)),(5)$$ where $a,b$ are real constants with $b=1$. By (5) we have $y^n=g_0(y)h_0(y)$. Consequently $g_0(y)=a^{\prime }{y}^s$, where $s\in {\mathbb{N}}^{\ast },s\le n$ and $a^{\prime }$ is a real constant, different from $0$. Put $c=-\frac{a^{\prime }}{a}$, we have $c\ne 0$. Plug $x=c{y}^s$ in (5), we obtain $$c^n{y}^{sn}+c{y}^{s+1}+y^n=0.(6)$$ + 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 $sn>n$ and $sn>s+1$. 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.