International Mathematical Olympiad Shortlisted Problems

Let $P(x)=a_n{x}^n+\cdots +a_0{x}^0$ with $a_n\ne 0$. Comparing the coefficients of $x^{n+1}$ on both sides gives $a_n(n-2m)(n-1)=0$, so $n=1$ or $n=2m$. If $n=1$, one easily verifies that $P(x)=x$ is a solution, while $P(x)=1$ is not. Since the given condition is linear in $P$, this means that the linear solutions are precisely $P(x)=tx$ for $t\in \mathbb{R}$. Now assume that $n=2m$. The polynomial $xP(x+1)-(x+1)P(x)=(n-1)a_n{x}^n+\cdots$ has degree $n$, and therefore it has at least one (possibly complex) root $r$. If $r\notin \{0,-1\}$, define $k=P(r)/r=P(r+1)/(r+1)$. If $r=0$, let $k=P(1)$. If $r=-1$, let $k=-P(-1)$. We now consider the polynomial $S(x)=P(x)-kx$. It also satisfies (1) because $P(x)$ and $kx$ satisfy it. Additionally, it has the useful property that $r$ and $r+1$ are roots. Let $A(x)=x^3-m{x}^2+1$ and $B(x)=x^3+m{x}^2+1$. Plugging in $x=s$ into (1) implies that: If $s-1$ and $s$ are roots of $S$ and $s$ is not a root of $A$, then $s+1$ is a root of $S$. If $s$ and $s+1$ are roots of $S$ and $s$ is not a root of $B$, then $s-1$ is a root of $S$. Let $a⩾0$ and $b⩾1$ be such that $r-a,r-a+1,\dots ,r,r+1,\dots ,r+b-1,r+b$ are roots of $S$, while $r-a-1$ and $r+b+1$ are not. The two statements above imply that $r-a$ is a root of $B$ and $r+b$ is a root of $A$. Since $r-a$ is a root of $B(x)$ and of $A(x+a+b)$, it is also a root of their greatest common divisor $C(x)$ as integer polynomials. If $C(x)$ was a non-trivial divisor of $B(x)$, then $B$ would have a rational root $\alpha$. Since the first and last coefficients of $B$ are $1,\alpha$ can only be 1 or -1 ; but $B(-1)=m>0$ and $B(1)=m+2>0$ since $n=2m$. Therefore $B(x)=A(x+a+b)$. Writing $c=a+b⩾1$ we compute $$0=A(x+c)-B(x)=(3c-2m)x^2+c(3c-2m)x+c^2(c-m).$$ Then we must have $3c-2m=c-m=0$, which gives $m=0$, a contradiction. We conclude that $f(x)=tx$ is the only solution.

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Alternative solution.

Multiplying (1) by $x$, we rewrite it as $$x\left(x^3-m{x}^2+1\right)P(x+1)+x\left(x^3+m{x}^2+1\right)P(x-1)=[(x+1)+(x-1)]\left(x^3-mx+1\right)P(x).$$ After regrouping, it becomes $$\begin{array}{c}\left(x^3-m{x}^2+1\right)Q(x)=\left(x^3+m{x}^2+1\right)Q(x-1),\end{array}\text{\hspace{1em}(2)}$$ where $Q(x)=xP(x+1)-(x+1)P(x)$. If $\mathrm{deg}P⩾2$ then $\mathrm{deg}Q=\mathrm{deg}P$, so $Q(x)$ has a finite multiset of complex roots, which we denote $R_Q$. Each root is taken with its multiplicity. Then the multiset of complex roots of $Q(x-1)$ is $R_Q+1=\{z+1:z\in R_Q\}$. Let $\{x_1,x_2,x_3\}$ and $\{y_1,y_2,y_3\}$ be the multisets of roots of the polynomials $A(x)=x^3-m{x}^2+1$ and $B(x)=x^3+m{x}^2+1$, respectively. From (2) we get the equality of multisets $$\{x_1,x_2,x_3\}\cup R_Q=\{y_1,y_2,y_3\}\cup (R_Q+1).$$ For every $r\in R_Q$, since $r+1$ is in the set of the right hand side, we must have $r+1\in R_Q$ or $r+1=x_i$ for some $i$. Similarly, since $r$ is in the set of the left hand side, either $r-1\in R_Q$ or $r=y_i$ for some $i$. This implies that, possibly after relabelling $y_1,y_2,y_3$, all the roots of (2) may be partitioned into three chains of the form $\{y_i,y_i+1,\dots ,y_i+k_i=x_i\}$ for $i=1,2,3$ and some integers $k_1,k_2,k_3⩾0$. Now we analyze the roots of the polynomial $A_a(x)=x^3+a{x}^2+1$. Using calculus or elementary methods, we find that the local extrema of $A_a(x)$ occur at $x=0$ and $x=-2a/3$; their values are $A_a(0)=1>0$ and $A_a(-2a/3)=1+4a^3/27$, which is positive for integers $a⩾-1$ and negative for integers $a⩽-2$. So when $a\in \mathbb{Z},A_a$ has three real roots if $a⩽-2$ and one if $a⩾-1$. Now, since $y_i-x_i\in \mathbb{Z}$ for $i=1,2,3$, the cubics $A_m$ and $A_{-m}$ must have the same number of real roots. The previous analysis then implies that $m=1$ or $m=-1$. Therefore the real root $\alpha$ of $A_1(x)=x^3+x^2+1$ and the real root $\beta$ of $A_{-1}(x)=x^3-x^2+1$ must differ by an integer. But this is impossible, because $A_1\left(-\frac{3}{2}\right)=-\frac{1}{8}$ and $A_1(-1)=1$ so $-1.5<\alpha <-1$, while $A_{-1}(-1)=-1$ and $A_{-1}\left(-\frac{1}{2}\right)=\frac{5}{8}$, so $-1<\beta <-0.5$. It follows that $\mathrm{deg}P⩽1$. Then, as shown in Solution 1, we conclude that the solutions are $P(x)=tx$ for all real numbers $t$.