Balkan Mathematical Olympiad

Assume that $f_n(x,y,z)$ divides $g_n(x,y,z)$, that is $$g_n(x,y,z)=f_n(x,y,z)h_n(x,y,z),$$ where $h_n$ is a polynomial in $x,y,z$ with integer coefficients. Considering $x=2,y=1,z=0$, it follows $2^{2n}-1$ must divide $(-2{)}^{5n}+2$. The last property is equivalent to $2^{2n}-1|2^{5n}+2(-1{)}^{5n}=2^{5n}+2(-1{)}^n=2^n(2^{4n}-1+1)+2(-1{)}^n=2^n(2^{4n}-1)+2^n+2(-1{)}^n$, hence $2^{2n}-1$ divides $2^n+2(-1{)}^n$. For $n\ge 2$ clearly we have $2^{2n}-1>2^n+2\ge 2^n+2(-1{)}^n$, therefore the only possibility is $n=1$. We will show that $f_1(x,y,z)$ divides $g_1(x,y,z)$. Denote $a=x-y,b=y-z,c=z-x$ and $S_k=a^k+b^k+c^k$ for $k=0,1,\dots$. Clearly, we have $a+b+c=0$, and from Newton's formulas it follows $$S_{k+3}=-(ab+bc+ca)S_{k+1}+abc{S}_k,k=0,1,\dots (1)$$ Also, from $a+b+c=0$ we obtain $a^2+b^2+c^2=-2{\sum }_{cyc}ab$, that is $$a^2+b^2+c^2-\sum _{cyc}ab=-3\sum _{cyc}ab.(2)$$ Using formulas (1) and (2) we have $$\begin{array}{rl}{g}_1(x,y,z)& =S_5=-(\sum _{cyc}ab)S_3+abc{S}_2=abc(S_2-3\sum _{cyc}ab)=\\ & =-5abc(\sum _{cyc}ab)=\frac{5}{2}abc{S}_2=5(x-y)(y-z)(z-x)f_1(x,y,z),\end{array}$$ hence the divisibility holds in this case.

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Alternative solution.

From the relation $g_n(x,y,z)=f_n(x,y,z)h_n(x,y,z)$, it follows that for every triple $(x,y,z)$ with the property $f_n(x,y,z)=0$ we have $g_n(x,y,z)=0$. We will consider two cases.

Case 1. If $n$ is even, we have $g_n(x,y,y)=2(x-y{)}^{5n}=f_n(x,y,y)h_n(x,y,y)$. This is equivalent to $f_n(x,y,y)=x^{2n}+2y^{2n}-2xy-y^2$ divides $2(x-y{)}^{5n}$, hence $x^{2n}+2y^{2n}-2xy-y^2=k(x-y{)}^t$, for some integer $k$ and some positive integer $t$. From the equality of the degrees it follows $t=2n$ and from the equality of the coefficients of $x^{2n}$, respectively $y^{2n}$, we obtain $k=1$ and $k=2$, not possible.

Case 2. If $n$ is odd and $n\ge 3$, then we have $f_n(-y,y,iy)=y^{2n}-y^2$ and $g_n(-y,y,iy)=y^{5n}[(-2{)}^{5n}+(1-i{)}^{5n}+(1+i{)}^{5n}]$. For $y^{2n-2}=1$ we have $f_n(-y,y,iy)=0$, hence $g_n(-y,y,iy)=0$. That is $(\frac{1-i}{2}{)}^{5n}+(\frac{1+i}{2}{)}^{5n}=1$. On the other hand, we have $$1=|(\frac{1-i}{2}{)}^{5n}+(\frac{1+i}{2}{)}^{5n}|=|2Re(\frac{1+i}{2}{)}^{5n}|\le 2|(\frac{1+i}{2}{)}^{5n}|=\frac{2}{(\sqrt{2}{)}^{5n}}<1,$$ contradiction. The only possibility is $n=1$ and we continue as in the last part of Solution 1.