Balkan Mathematical Olympiad Shortlisted Problems

Suppose that $n$ is a solution to the problem. The polynomial $$P(x)=x^n-4x^2+2x+1$$ clearly has a root at $1$. Therefore we may write $P(x)=(x-1)Q(x)$ for some polynomial $Q$. Since $P(x)\ge 0$ for $x>0$, the polynomial $Q$ changes sign at $1$ and so $Q(1)=0$. Calculating $$Q(x)=\frac{P(x)}{x-1}=\frac{x^n-1}{x-1}-\frac{4x^2-2x-2}{x-1}=(1+x+\cdots +x^{n-1})-(4x+2)$$ we see that $Q(1)=n-6$ and hence $n=6$.

Conversely, let $n=6$. Note that $Q(x)=(x-1)R(x)$, where $R$ is a polynomial given by: $$R(x)=x^4+2x^3+3x^2+4x+1$$ Since $P(x)=(x-1{)}^{2}R(x)$ and $R(x)\ge 0$ for $x>0$, we indeed have that $P(x)\ge 0$ for $x>0$. It follows that $n=6$ is a solution, and thus the only solution, to the problem.

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Alternative solution.

Let $n$ be a solution to the problem. Consider the function $P:(0,+\infty )\to \mathbb{R}$, defined by $P(x)=x^n-4x^2+2x+1$ for every $x>0$. Since $P(x)\ge 0$ for every $x>0$ and $P(1)=0$, it follows that the function $P$ attains its minimum at $x=1$. Since this is also a local minimum and $P$ is differentiable, we have $P^{\prime }(1)=0$. Therefore, $P^{\prime }(1)=n-8+2=0$. Hence, $n=6$ is the only possible value that can be a solution to the problem. On the other hand, for $x>0$, using the AM-GM inequality, we have $$x^6+x+x+1\ge 4\sqrt[4]{x^6\cdot x\cdot x\cdot 1}=4x^2$$ This proves that $n=6$ is the only solution to the problem.

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Alternative solution.

We shall prove that the only solution is $n=6$. The given condition is equivalent to $P(x)\ge 0$ for all $x>0$, where $P(x)=x^n-4x^2+2x+1$. Using elementary transformations, for $n>2$, we have: $$\begin{array}{rl}P(x)& =x^n-1-4x^2+2x+2=(x-1)(x^{n-1}+x^{n-2}+\cdots +x+1)-(x-1)(4x+2)=\\ & =(x-1)(x^{n-1}+x^{n-2}+\cdots +x^2-3x-1)=(x-1)(x^{n-1}-1+x^{n-2}-1+\cdots +x^2-1-3x+3+n-6)=\\ & =(x-1)\left((x-1)(x^{n-2}+2x^{n-3}+3x^{n-4}+\cdots +(n-2)x+n-5)+n-6\right)=\\ & =(x-1{)}^2(x^{n-2}+2x^{n-3}+3x^{n-4}+\cdots +(n-2)x+n-5)+(n-6)(x-1).\end{array}\text{\hspace{1em}(P)}$$ From (P), since for $n=6$ and $x>0$ we have $Q(x)>0$, it follows that $n=6$ is one solution. Since the polynomial $Q(x)$ has all positive coefficients (except possibly the constant term), we have $Q(1-\frac{1}{3n})<Q(1)=\frac{n^2-n-8}{2}$, and based on (P), for $n\ge 7$, it holds that $$P\left(1-\frac{1}{3n}\right)=\frac{Q\left(1-\frac{1}{3n}\right)}{9n^2}-\frac{n-6}{3n}<\frac{Q(1)}{9n^2}-\frac{n-6}{3n}=\frac{5n(7-n)-8}{18n^2}<0.$$ This shows that the numbers $n\ge 7$ are not solutions to the problem. On the other hand, for $1\le n\le 5$, we have $$P(1.1)\le 1.1^5-4\cdot 1.1^2+2\cdot 1.1+1=-0.02949<0,$$ so the numbers $1\le n\le 5$ are also not solutions.

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Alternative solution.

Let $P(x)=x^n+2x+1-4x^2$, we want to check if $P(x)\ge 0$ holds for all $x>0$. If $n=1$ substituting $x=2$ yields $P(x)=-11<0$, meaning that $n=1$ cannot be our solution. From now on, assume that $n\ge 2$, and substitute $x=1+t$. Using the binomial theorem we obtain that $P(1+t)=(1+t{)}^n+2(1+t)-4(1+t{)}^2=(n-6)t-4t^2+{\sum }_{k=2}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)t^k\le (n-6)t+{\sum }_{k=2}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)t^k$. It is trivially true for $k\ge 2$ that $\left(\genfrac{}{}{0ex}{}{n}{k}\right)t^k<n^k|t{|}^k$, which implies $P(1+t)<(n-6)t+(nt{)}^2{\sum }_{k=0}^{n-2}(n|t|{)}^k$. Utilizing the identity that for $x\ne 1$ we have that ${\sum }_{k=0}^m{x}^k=\frac{1-x^{m+1}}{1-x}$ we obtain that $P(1+t)<(n-6)t+(nt{)}^2\frac{1-n^{n-1}|t{|}^{n-1}}{1-n|t|}=t(n-6+t{n}^2\frac{1-n^{n-1}|t{|}^{n-1}}{1-n|t|})$. If $n<6$, then let $t=\frac{1}{n^3}$, plugging this in we obtain $P\left(1+\frac{1}{n^3}\right)<\frac{1}{n^3}\left(n-6+\frac{1-\frac{1}{n^{2}n-2}}{n-\frac{1}{n}}\right)$. Now since $n<6$ and $n\in \mathbb{N}$, we have that $n-6\le -1$. On the other hand, we know that $1-\frac{1}{n^{2}n-2}<1<\frac{3}{2}\le n-\frac{1}{n}$ for $n\ge 2$, implying that $\frac{1-\frac{1}{n^{2}n-2}}{n-\frac{1}{n}}<1$. Therefore the expression in the parenthesis is negative, while $\frac{1}{n^3}>0$, meaning that the whole expression is negative, or $P\left(1+\frac{1}{n^3}\right)<0$, meaning that we found a negative value for $P(x)$ for $x=1+\frac{1}{n^3}>0$. Therefore no $n\in \{2,3,4,5\}$ can be a solution. If $n>6$, then let $t=-\frac{1}{n^3}$. Similarly to the case $n<6$, we obtain that $P(1-\frac{1}{n^3})<-\frac{1}{n^3}\left(n-6+\frac{1-\frac{1}{n^{2}n-2}}{n-\frac{1}{n}}\right)$. Note that $n-6\ge 1$, and it still holds that $\frac{1-\frac{1}{n^{2}n-2}}{n-\frac{1}{n}}<1$ using the same arguments. Therefore, the expression in the parenthesis is positive, while $-\frac{1}{n^3}<0$, meaning that $P\left(1-\frac{1}{n^3}\right)<0$, and since for $n>6$ we have $1-\frac{1}{n^3}>0$, we again found a negative point of our polynomial for some positive input. Therefore, the only case left is when $n=6$. For this take $x^6+x+x+1\ge 4x^2$ using AM-GM. Therefore, the only solution is $n=6$.