The 65th IMO China National Team Selection Test

Proof. We will sequentially prove the following conclusions:

(1) The equation (5) has 99 roots with modulus equal to 1.

Clearly, $x=1$ is a root of the equation (5). Consider the equation $$\frac{x^{101}-n{x}^{100}+nx-1}{x-1}=0,$$ i.e., $$f(x)=x^{100}-(n-1)\sum _{j=1}^{99}{x}^j+1=0.$$ Consider all 100th roots of unity $\omega$, it is easy to see that $f(\omega )=n+1$. Notice that $\frac{f(x)}{x^{50}}$ can be expressed as $g(x+\frac{1}{x})$, where $g(x)\in \mathbb{Z}[x]$. Therefore, $$g\left(2\cos \frac{k\pi }{50}\right)={\left(e^{ik\pi /50}\right)}^{50}f\left(e^{ik\pi /50}\right)=(n+1)(-1{)}^k,k=0,1,2,\dots ,49.$$ This shows that $g(2\cos \theta )$ has a root in the interval $(\frac{k\pi }{50},\frac{(k+1)\pi }{50})$ with $\theta ={\theta }_k$ ($k=0,1,2,\dots ,48$). Therefore, $g(x)$ has 49 pairs of conjugate complex roots $\cos {\theta }_k\pm i\sin {\theta }_k$. Thus, $f(x)$ has 98 roots with modulus equal to 1, and hence the original equation (5) has 99 roots with modulus equal to 1.

(2) $f(x)$ has a root $\alpha$ greater than 1.

This is because $f(1)=2-99(n-1)<0$, which follows from the intermediate value theorem. Combining $f(0)=1>0$ and the fact that the product of all roots of $f(x)$ is 1, we know that besides the 99 roots with modulus equal to 1, the remaining two roots are $\alpha$ and $\frac{1}{\alpha }$.

(3) The 99 roots of $f(x)$ with modulus equal to 1 are not all roots of unity.

If any of these roots are roots of unity, the cyclotomic polynomial corresponding to that root divides $f(x)$. Therefore, if they are all roots of unity, this means $(x-\alpha )(x-\frac{1}{\alpha })=x^2-Ax+1$, where $A$ is a positive integer. If $A\ge n+1$, then $\alpha \ge n$, thus ${\alpha }^{101}-n{\alpha }^{100}+n\alpha -1>0$, which is a contradiction! If $A\le n$, then using $(\alpha +1)({\alpha }^2-A\alpha +1)={\alpha }^3-(A-1){\alpha }^2-(A-1)\alpha +1$, we get $${\alpha }^3<(A-1){\alpha }^2+(A-1)\alpha \le (n-1){\alpha }^2+(n-1)\alpha .$$ Therefore, $${\alpha }^{100}-(n-1)({\alpha }^{99}+{\alpha }^{98}+\cdots +\alpha +1)<-(n+1)({\alpha }^{97}+\cdots +\alpha )+1<0,$$ which is a contradiction! Thus, the assumption is not valid, meaning that the roots of (5) are not all roots of unity.

Since the powers of the roots of unity are finite in number, and using Newton's identities, we know that the sum of the powers of all roots of the original equation are integers. To prove the original problem, we need to show the following conclusion:

If $z_1,z_2,\dots ,z_k$ are complex numbers with modulus 1 but not roots of unity, then the fractional parts of $S_r={\sum }_{j=1}^k(z_j^r+{\bar{z}}_j^r)$ are dense in $(0,1)$. That is: If ${\lambda }_1,\dots ,{\lambda }_k$ are irrational numbers, then the fractional parts of $T_r={\sum }_{j=1}^k\cos (2r{\lambda }_j\pi )$ are dense in $(0,1)$.

Consider a sufficiently large positive integer $N$, and define the sequence $$X_r=(\lfloor N\{r{\lambda }_1\}\rfloor ,\dots ,\lfloor N\{r{\lambda }_k\}\rfloor )(r=1,2,\dots ,N^k+1).$$ By the pigeonhole principle, there exist $r_1<r_2$ such that $X_{r_1}=X_{r_2}$. This means that for $s=r_2-r_1$, $\{s{\lambda }_j\}$ is either less than $\frac{1}{N}$ or greater than $1-\frac{1}{N}$, which implies $\cos (2s{\lambda }_j\pi )>\cos \frac{2\pi }{N}$. Therefore, $T_s>k\cos \frac{2\pi }{N}$. On the other hand, since $\{s{\lambda }_1\}\ne 0$, there exists a positive integer $t$ such that $|t\{s{\lambda }_1\}-\frac{1}{2}|<\frac{1}{N}$, hence $\cos (2st{\lambda }_1\pi )<-\cos \frac{2\pi }{N}$. Therefore, $T_{st}<(k-1)-\cos \frac{2\pi }{N}$. But we know, $$\begin{array}{rl}|T_{(d+1)s}-T_{ds}|& =\left|\sum _{j=1}^k(\cos (2(d+1)s{\lambda }_j\pi )-\cos (2ds{\lambda }_j\pi ))\right|\\ & =\left|\sum _{j=1}^{k}2\sin (s{\lambda }_j\pi )\sin ((2d+1)s{\lambda }_j\pi )\right|\\ & \le \sum _{j=1}^k|2\sin (s{\lambda }_j\pi )|<2k\sin \frac{2\pi }{N},\end{array}$$ therefore, by taking $N$ sufficiently large, using the discrete intermediate value theorem, we can see that $T_r$ is dense in $(k-\frac{3}{2},k-\frac{1}{2})$.

The desired conclusion is thus proven. □