Let $m \neq 0 $ be an integer. Find all polynomials $P(x) $ with real coefficients such that \[ (x^3 - mx^2 +1 ) P(x+1) + (x^3+mx^2+1) P(x-1) =2(x^3 - mx +1 ) P(x) \] for all real number $x$.
Problem
Source: IMO Shortlist 2013, Algebra 6
Tags: algebra, polynomial, IMO Shortlist
shmm
12.07.2014 13:05
Does anybody have solutions?
xxp2000
12.07.2014 17:06
Obviously $P(x)=cx$ is a solution. We will show there are no other solutions. Suppose $P$ is a solution with degree $n>1$. We can rewrite it as $(x^3+1)(P(x+1)+P(x-1)-2P(x))=mx(xP(x+1)-xP(x-1)-2P(x))$. Compare the coefficients of $x^{n+1}$ on both sides, we get $n=2m$. So $m>0$. We can verify there is no solution when $m=1,n=2$. Now consider $m\ge2$. Let $a,b$ be the largest real root of $x^3-mx^2+1$ and $x^3+mx^2+1$, respectively. We can verify $a\in (m-1,m),b\in (-m-1,-m)$. Also $a-2m$ can not be a root of $x^3+mx^2+1$. If $P$ is a solution, so is $P(x)-\frac{P(a)}ax$. So we can assume $P(a)=0$ wlog. Let $x=a,a-1,\cdots,a-2m+1$, we get $P(a-i)=0,i=1,2,\cdots,2m$. So $P$ has at least $2m+1=n+1$ distinct roots. Absurd!
va2010
11.10.2016 00:20
The above solution is quite clever, but mine is perhaps easier to find. The key step is as follows. If $Q(x) = xP(x+1) - (x+1)P(x)$, it must be true that \[ (x^3 - mx^2 + 1)Q(x) = (x^3 + mx^2 + 1)Q(x - 1) \]. This can be shown by an expansion, for when we multiply this out it becomes the original relation multiplied by $x$. There are many finishes from here, but this is the one I think is nicest. By Descarte's Rule of Signs, if $a > 0$, then $x^3 + ax^2 + 1$ has only one real root. Moreover, if $a \le -2$, $R(x) = x^3 + ax^2 + 1$ must have three real roots, as $R(-a)$ is positive, $R(1)$ is negative, $R(0)$ is positive, and $R(a)$ is negative, so there are at least two real roots and since the number of nonreal roots it even all three roots are real. Hence, for $|m| \neq 1$ it must be true that $Q$ is identitcally zero. For $|m| = 1$, any root of $x^3 + mx^2 + 1$ must be root of $Q(x)$, as it cannot be a root of $x^3 - mx^2 + 1$. Any root of $x^3 - mx^2 + 1$ also must be a root of $Q(x-1)$, which means that the roots of $x^3 - x^2 + 1$ differ from the roots of $x^3 + x^2 + 1$ by one each, which we can easily see to be wrong (just plug in $x + 1$ and $x - 1$ into $x^3 - x^2 + 1$ and compare with $x^3 + x^2 + 1$). Therefore, in both cases we conclude that $Q$ has infinite roots and is therefore identically zero. If $xP(x+1) = (x+1)P(x)$, it must be true that $P(x+1)$ has $-1$ as a root, so $P(x) = xQ(x)$, and $Q(x) = Q(x+1)$ for all $x$. If $Q$ has a finite, nonzero number of roots, let $r$ be the largest root, but then $r + 1$ is a root. Therefore, the only possibilities are for $Q$ to either be nonzero or zero, so it must be constant and $P(x) = cx$ for all $x$.
v_Enhance
30.03.2017 01:48
Linear polynomials $P(x) = cx$ work and are the only solution. To make sense of what is going on, we move all the terms involving $m$ to one side of the equation: \[ (x^3+1) \left( P(x+1) - 2P(x) + P(x-1) \right) = mx \cdot \left( xP(x+1) - 2P(x) - xP(x-1) \right). \]The left-hand side is the second Mahler difference of $P$. Now, if we define \begin{alignat*}{4} Q(x) &= xP(x+1) &-(x+1) P(x) & Q(x-1) &= & (x-1)P(x) &- xP(x-1). \end{alignat*} then $Q(x) \pm Q(x-1)$ generate the relevant expressions: we obtain \[ \frac{x^3+1}{x} \left( Q(x) - Q(x-1) \right) = mx(Q(x) + Q(x-1)) \]or \[ \boxed{\left( x^3 - mx^2 + 1 \right) Q(x) = \left( x^3 + mx^2 + 1 \right) Q(x-1)}. \]This is the main work in the problem, and the rest is bookkeeping. We contend that $Q$ is the zero polynomial, which will directly imply that $P$ is linear. To do this we need to just show that: Lemma: There exists a complex root $\alpha$ of $R^-(x) \overset{\text{def}}{=} x^3-mx^2+1$ such that $\alpha+n$ is not a root of $R^+(x) \overset{\text{def}}{=} x^3+mx^2+1$ for any $n \in {\mathbb Z}$. Proof. If $m=2$ take $\alpha = 1$. Otherwise if $m \neq 0, 2$ then $R^-$ is irreducible (hence squarefree). Then we claim that we may select any irrational root $\alpha$. Indeed, if not then the roots $\alpha$, $\beta$, $\gamma$ are Galois conjugates (since $R^-$ irreducible), so if $\alpha+n$ is a root of $R^+$, so are $\beta+n$ and $\gamma+n$. Now $\alpha+\beta+\gamma = m$ and $\alpha\beta + \beta\gamma + \gamma\alpha = n$. On the other hand $-m = \sum \alpha+n \implies 2m = 3n$, but now \begin{align*} -1 = (\alpha+n)(\beta+n)(\gamma+n) &= -1 + n \cdot 0 + n^2 (-m) + n^3 \\ \implies 0 & n^2(m+n). \end{align*}Together with $2m = 3n$ this forces $m = n = 0$, so $R^+ = R^-$, which is absurd. $\blacksquare$ Then $\alpha$ must be a root of $Q(x-1)$, so $\alpha+1$ is a root of $Q$, and by induction $\alpha+n$ is a root of $Q$ for all $n$, hence $Q$ is the zero polynomial. This implies $P$ is linear as desired. Remark: If $m = 0$ was allowed, then we would get $Q$ constant instead of $Q \equiv 0$; this would mean linear solutions $P$ were valid.
jeff10
04.09.2019 05:58
The most natural thing to start doing is verifying which constant polynomials work. I didn't verify linear polynomials until later when working on the problem, but I'll do that here anyways. Through writing $P(x)=ax+b$, one finds that $P(x)=cx$ work for all real numbers $c$.
After seeing the mismatching terms $-mx^2$, $mx^2$, and $-mx$, it is tempting to conjecture that the polynomials above are the only valid solutions for $P(x)$, but this is what we must prove or disprove.
Assume that the degree of $P(x)$ is at least $2$. If we write the polynomial $P(x)$ as $a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$, where $a_n \ne 0$, and plug in the respective forms into the equation condition, we can start comparing coefficients. We do not discover anything useful from comparing the coefficients of $x^{n+3}$ or $x^{n+2}$. But if we compare coefficients for $x^{n+1}$, we get the relation $$(\binom{n}{2}a_n+(n-1)a_{n-1}+a_{n-2})-m(na_n+a_{n-1})+(\binom{n}{2}a_n-(n-1)a_{n-1}+a_{n-2})+m(-na_n+a_{n-1})=2(a_{n-2}-ma_n)$$which, after multiplication by $2$, reduces to $$n(n-1)a_n=2m(n-1)a_n$$Since $a_n \ne 0$ and $n-1 \ge 1 > 0$, we can cancel out those terms to get what seems to be a crucial relation: $n=2m$.
If we plug in $x=0$ into the original equation, we get $P(1)+P(-1)=2P(0)$. If $P(x)=ax^2+bx+c$ was a quadratic, we would get the relation $(a+b+c)+(a-b+c)=2c$, or $a=0$, which means $P(x)$ cannot be a quadratic.
Combining these two observations, we observe that $n=2m > 2$. Since $m$ is an integer, we must have $m \ge 2$ and $n \ge 4$.
With this information, it seems like we want to get more out of what $n$ of $m$ could be, making the next natural step to compare more coefficients, since we have that $P(x)$ must be at least a quartic. Turns out that it does not do much for us.
Now I want to focus on the original polynomials $x^3-mx^2+1$, $x^3+mx^2+1$, and $x^3-mx+1$, since they are polynomial factors appended to variants of the unknown polynomial $P(x)$. It would be nice to be able to plug in roots of these expressions to reduce the problem into one comparing only two general terms, but that only happens when any of the polynomials can factor. Since these are cubics, we need them to have a rational root. The Rational Root Theorem guarantees that the only possibilities to check are $x=1$ and $x=-1$, but this only gives possibilities of $m=-2$ and $m=2$, the former of which is impossible. So let's attempt the latter exclusively.
Plugging in $x=1$, we get $P(0)=0$. This allows us to write $P(x)$ as $xQ(x)$, where $Q(x)$ is a polynomial with real coefficients. Rewriting the given condition, we get that $$[A(x)-B(x)]Q(x+1)+[A(x)+B(x)]Q(x-1)=2A(x)Q(x)$$where $A(x)=x^4-2x^2+x$ and $B(x)=x^3-1$. Isolating $A(x)$ and $B(x)$ onto different sides, we have the rearranged equation $$A(x)[Q(x+1)-2Q(x)+Q(x-1)]=B(x)[Q(x+1)-Q(x-1)]$$Canceling $x-1$ from both sides and expressing $D(x)=Q(x+1)-Q(x)$, we find that $$x(x^2+x-1)[D(x)-D(x-1)]=(x^2+x+1)[D(x)+D(x-1)]$$Now we isolate $D(x)$ and $D(x-1)$ onto opposite sides to get the equation $$(x^3-2x-1)D(x)=(x+1)(x^2-x-1)D(x)=(x^3+2x^2+1)D(x-1)$$Because we know nothing about the polynomial $D(x)$ other than it has real coefficients, it only makes sense to look at roots of expressions in the equations above. Let $\varphi$ be a root of $x^2-x-1$. If $\varphi$ was a root of $x^3+2x^2+1$, then $x^3+2x^2+1$ can be factored into a quadratic expression and a linear expression, both with integer coefficients. This would force $\varphi$ to be an integer, a contradiction. Hence, $\varphi$ must be a root of $D(x-1)$. This means that $D(\varphi-1)=0$, so that $\varphi-1$ is a root of $D(x)$. We know what solutions to quadratic equations look like, and $\varphi-1$ is still in that form. So if $\varphi$ is a root of $x^3+2x^2+1$, then it would be rational, a contradiction. Hence, $\varphi-1$ is a root of $D(x-1)$, which would mean $\varphi-2$ is a root of $D(x)$. We notice that we can inductively continue this pattern to show that $\varphi-z$ is a root of $D(x)$ for all positive integers $z$. It follows that $D(x)$ has infinitely many roots, forcing $D(x)=0$.
Given $D(x)=0$, we are now forced to have $Q(x)$ be a constant $c$, which would mean that $P(x)$ can only have the form $cx$ for some real number $c$. These are the polynomials described in general at the beginning of our solution.
What if $m > 2$? This time, we no longer have the luxury of $x=1$ being a nice root to give us $P(0)=0$. However, since showing that a polynomial $P(x)$ is satisfied for infinitely many values of $x$ in the given equation is sufficient, I might still want to somehow work with this. Seeing that we had polynomials lined up so nicely in our work for $m=2$, I want to go ahead and write $P(x)=xQ(x)$ anyways. Now $Q(x)$ is a rational function (where the denominator is $1$ or $x$) instead of a polynomial. As worrying as $x=-1, 0, 1$ can be, if we want to satisfy our given equation for an infinite number of values $x$, these extraneous cases will not matter.
Now we set $A(x)=x^4-2x^2+x$ and $B(x)=(m-1)x^3-1$. This will give us the same equation from our work on $m=2$ before: $$[A(x)-B(x)]Q(x+1)+[A(x)+B(x)]Q(x-1)=2A(x)Q(x)$$which we can rearrange as $$A(x)[Q(x+1)-2Q(x)+Q(x-1)]=B(x)[Q(x+1)-Q(x-1)]$$Writing $D(x)=Q(x+1)-Q(x)$ (which is a rational function with denominator a factor of $x(x+1)$), we get $$A(x)[D(x)-D(x-1)]=B(x)[D(x)+D(x-1)]$$We use the same method as before on this equation to get $$[A(x)-B(x)]D(x)=[A(x)+B(x)]D(x-1)$$If we had written out the expansions for the given equation when we express $P(x)=xQ(x)$, the expression $A(x)-B(x)$ and $A(x)+B(x)$ will look familiar. Indeed, they are the polynomials $(x^3-mx^2+1)(x+1)$ and $(x^3+mx^2+1)(x-1)$, respectively. So then why did I go through the trouble of writing them broken as $A(x)$ and $B(x)$ and with another difference condition $D(x)=Q(x+1)-Q(x)$? My response would be that the $D(x)$ functions give rise to an equation involving equality between a product of polynomials. When we have such a product with no adequately restrictive conditions, the best thing we can do is relate roots of expressions, and in this instance, relating roots is quite convenient.
After a slight bit of work, we find that the greatest common divisor of the polynomials $x^3-mx^2+1$ and $x^3+mx^2+1$ in the ring $\mathbb{C}[x]$ is $1$, as they have no complex roots in common, and the Fundamental Theorem of Algebra guarantees that all the roots of both of these polynomials must be complex.
For simplicity, I will denote $A(x)-B(x)$ by $G(x)$ and $A(x)+B(x)$ by $H(x)$, so that the equation is now $$G(x)D(x)=H(x)D(x-1)$$Let's take a root $r$ of $G(x)$ that is not $-1$ (so a root of $x^3-mx^2+1$). Because it is not a root of $H(x)$, it must satisfy $D(r-1)=0$. We don't have the issue of the denominator being zero here because neither $x^3-mx^2+1$ nor $x^3+mx^2+1$ can have rational roots. This means we have $r-1$ being a zero of $D(x)$. We would like for $r-1$ not to be a root of $H(x)$ so that we can conclude that $r-2$ is a zero of $D(x)$, and we would like $r-2$ to not be a root of $H(x)$ so that we can conclude that $r-3$ is a zero of $D(x)$, and so forth. The issue here is that we do not know whether between $x^3-mx^2+1$ and $x^3+mx^2+1$ there exists a root of one polynomial that does not differ from any root of the other polynomial by an integer. We will remedy this issue below.
For every root $y$ of either $x^3-mx^2+1$, the minimal polynomial of $y$ must be $x^3-mx^2+1$, or else $x^3-mx^2+1$ can be factored in $\mathbb{Z}[x]$. The same holds true for $x^3+mx^2+1$.
If It is easy to see that if $y_1$, $y_2$, $y_3$ are the roots of a cubic, then if $z$ is any integer, then the Vieta expressions $(y_1+z)+(y_2+z)+(y_3+z)$, $(y_1+z)(y_2+z)+(y_2+z)(y_3+z)+(y_3+z)(y_1+z)$, and $(y_1+z)(y_2+z)(y_3+z)$ will all be integers. Combined with the fact that $y_i+z$ is never rational, we get that if $J(x)=x^3-mx^2+1$ is the minimal polynomial of the complex number $y$, then $J(x+z)$ is the minimal polynomial for the complex number $y+z$. In other words, $x^3+mx^2+1=J(x+z)$ is forced. Rewriting $z$ as $-z$, this will force $-(y_1-z)(y_2-z)(y_3-z)=(z-y_1)(z-y_2)(z-y_3)$ to equal $1$. Then what we are saying is that $J(z)=1$ for some integer $z$, which is the same as saying $z^3-mz^2+1=1$ for some integer $z$. That is, $z^3-mz^2=z^2(z-m)=0$. We do not care about $z=0$, so the only possibility here is $z=m$. However, if we do Vieta's on the sum of the roots, we will notice that if the $3$ roots in question are shifted by $m$, then the sum should be shifted by $3m$, but we see a shift by $2m$ when looking between $x^3-mx^2+1$ and $x^3+mx^2+1$ (I did not bother to look at sign since that is not a significant factor when the magnitudes of change differ).
It follows that it is impossible for any root of $x^3-mx^2+1$ to differ from any root of $x^3+mx^2+1$ by an integer. This allows us to perform the inductive procedure we were working with before, which will produce an infinite number of zeros of $D(x)$, none of which are rational. If we write $D(x)$ as a quotient of polynomials $\frac{U(x)}{V(x)}$, we will have an infinite number of roots for the polynomial $U(x)$, forcing $U(x)=0$. In turn, this forces $D(x)=0$, which forces $Q(x)$ to be a constant, which forces $P(x)=cx$ for some real number $c$.
In conclusion, for all integers $m \ne 0$, the only polynomials which satisfy the given equation are those of the form $\boxed{P(x)=cx}$, where $c$ is any real number.
CANBANKAN
07.06.2020 07:52
The only functions are $f(x)=cx$ for some $c\in \mathbb{R}$. It's clear that they work. Now I claim these are the only ones.
Let $n=deg (P)$, $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$. First note comparing the $x^{n+1}$ coefficient yields $n(n-1)=2mn-2m$, transposing to $n^2-(2m+1)n+2m=0$, $\therefore n\in \{1,2m\}$.
If $n=1$, let $P(x)=ax+b$. We have $$(x^3-mx^2+1)(ax+b+a)+(x^3+mx^2+1)(ax+b-a)=2(x^3-mx+1)(ax+b)$$
$$(ax+b)(2x^3-2mx+2-x^3+mx^2-1-x^3-mx^2-1)$$
$$-a(x^3-mx^2+1)+a(x^3+mx^2+1)=0$$
Which simplifies to $b=0$.
Assume $n=2m$. Let $Q(x)=xP(x+1)-(x+1)P(x)$. We can see $(x^3+mx^2+1)Q(x-1)=(x^3-mx^2+1)Q(x)$. Assume $Q(x)$ is not the zero polynomial.
I first claim $m>1$. Note $gcd(x^3+mx^2+1,x^3-mx^2+1)=gcd(2mx^2,x^3-mx^2+1)=1$ so $x^3+mx^2+1|Q(x)$, so $\deg(Q)=\deg(P)=2m\ge 3$, so $m\ge 2$.
We first analyze the number of real roots of $f(x)=x^3-mx^2+1$. Its derivative is $3x^2-2mx$ so $f'(x)$ changes sign at $x=0,\frac{2m}{3}$. At $x=0$, $f(x)=1$, and at $x=\frac{2m}{3}$, $f(x)=1-\frac{4m^3}{27}$. Since $m\ge 2, 1-\frac{4m^3}{27}<0$, $f(x)$ has three roots. Meanwhile, $x^3+mx^2+1$ has one real root.
Let $x$ be the number of real roots of $Q(x)$, counting multiplicity. It is also the number of real roots of $Q(x-1)$. However, $x^3-mx^2+1$ has three real roots while $x^3+mx^2+1$ has one, contradiction.
TheUltimate123
17.10.2020 01:25
Solved with nukelauncher. The linear \(P\) that work are \(P(x)\equiv cx\). We will show these are the only solutions. We can easily verify the following via substitution: for any root \(r\) of \(x^3-mx^2+1\), we have \(rP(r-1)=(r-1)P(r)\); and for any root \(r\) of \(x^3+mx^2+1\), we have \(rP(r+1)=(r+1)P(r)\). Hence we consider the polynomial \[Q(x):=(x-1)P(x)-xP(x-1),\]with the properties that \(Q(x+1)+Q(x)=P(x+1)-2P(x)-xP(x-1)\) and \(Q(x+1)-Q(x)=x(P(x+1)-2P(x)+P(x-1))\). After some brief computation, the given functional equation rewrites as \begin{align*} mx\big(xP(x+1)-2P(x)-xP(x-1)\big)&=\left(x^3+1\right)\big(P(x+1)-2P(x)+P(x-1)\big)\\ mx\cdot\big(Q(x+1)+Q(x)\big)&=\left(x^3+1\right)\cdot\big(Q(x+1)-Q(x)\big)\\ Q(x+1)\cdot\left(x^3-mx^2+1\right)&=Q(x)\cdot\left(x^3+mx^2+1\right).\tag{\(*\)} \end{align*} Claim: There is a root \(r\) of \(x^3-mx^2+1\) such that no element of \(r+\mathbb Z\) is a root of \(x^3+mx^2+1\). Proof. For \(m=2\), take \(r=1\); then \(x^3+2x^2+1\) has no integer root, as needed. Otherwise, by rational root theorem, \(x^3-mx^2+1\) will have three irrational roots \(r\), \(s\), \(t\); in particular, \(r\), \(s\), \(t\) are Galois conjugates. Then for any integer \(a\), the numbers \(r+a\), \(s+a\), \(t+a\) are also Galois conjugates, so is \(r+a\) is a root of \(x^3+mx^2+1\), then so are \(s+a\), \(t+a\). It follows that \begin{align*} x^3-mx^2+1&=(x-r)(x-s)(x-t)\\ x^3+mx^2+1&=(x-r-a)(x-s-a)(x-t-a). \end{align*}By comparing the quadratic terms, we have \(r+s+t=m\) and \(r+s+t+3a=-m\), so \(3a=-2m\). But by comparing the constant terms, we have \[-1=(r+a)(s+a)(t+a)=-P(-a)=a^3+ma^2-1,\]so \(a=0\) or \(a=m\). Either way, combining with \(3a=-2m\) gives \(a=m=0\), contradiction. \(\blacksquare\) Claim: \(Q\equiv0\). Proof. Assume for contradiction \(Q\not\equiv0\), i.e.\ \(Q\) has finitely many roots. Take \(s\) a root of \(Q(x)\). Then I claim there exists \(a\in\mathbb Z\) such that \(s+a\) is a root of \(x^3-mx^2+1\); otherwise, we will show by induction on \(i\ge0\) that \(Q(s+i)=0\). Indeed, if \(Q(s+i)=0\), then since \(s+i\) is not a root of \(x^3-mx^2+1\), by \(x=s+i\) in \((*)\), we have \(Q(s+i+1)=0\) as well. I claim there exists \(b\in\mathbb Z\) such that \(s-b\) is a root of \(x^3+mx^2+1\); otherwise, we will show by induction on \(i\ge0\) that \(Q(s-i)=0\). Indeed, if \(Q(s-i)=0\), then since \(s-i\) is not a root of \(x^3+mx^2+1\), by \(x=s-i-1\) in \((*)\), we have \(Q(s-i-1)=0\) as well. The above two conditions in tandem contradict the first claim. \(\blacksquare\) Therefore, \((x-1)P(x)\equiv xP(x-1)\); that is, \[\frac{P(x)}x=\frac{P(x-1)}{x-1}\]for all \(x\notin\{0,1\}\), so \(P\) is of the form \(P(x)\equiv cx\) for some \(c\).
yayups
03.02.2021 09:32
We will treat the statement as a formal polynomial identity, as it is true for all real values $x$, and in particular $x$ will be treated as a formal variable. Let $R(x) = xP(x+1) - (x+1)P(x)$. It is easy to verify that the given identity is equivalent to \[(x^3-mx^2+1)R(x) = (x^3+mx^2+1)R(x-1).\quad\quad(\star)\]Note that \[\gcd(x^3-mx^2+1,x^3+mx^2+1) = \gcd(x^3-mx^2+1,2mx^2)=1,\]so we must have $x^3-mx^2+1\mid R(x-1)$ and $x^3+mx^2+1\mid R(x)$. Suppose that $R$ is not constant. Now, suppose that $z$ is a (potentially complex) root of $R$ such that $z^3+mz^2+1\ne 0$. Then, $(\star)$ implies that $z-1$ is also a root of $R$. Furthermore, if $z^3-mz^2+1\ne 0$, then $(\star)$ implies that $z+1$ is also a root of $R$. Thus, there exist nonnegative integers $m$ and $n$ such that $z-m$ is a root of $x^3+mx^2+1$ and $z+n$ is a root of $x^3-mx^2+1$, since $R$ has only finitely many roots. Therefore, there is some complex number $r$ and integer $t$ such that \[r^3-mr^2+1=0\]and \[(r+t)^2+m(r+t)^2+1=0.\]It is easy to see that these two imply that there is some quadratic equation with rational coefficients that $r$ is a root of, which is a contradiction as $x^3-mx^2+1$ is irreducible (has no rational roots), so it can't have a smaller degree rational polynomial which it is a root of. This shows that $R$ is constant, and then $(\star)$ implies that in fact $R\equiv 0$. Thus, \[\frac{P(x+1)}{x+1} = \frac{P(x)}{x}\]for all $x$. This implies that $P(x)=cx$ for some real constant $c$, which works. This completes the solution. Remark: We'll provide some motivation for the magical $R(x)$ substitution. The equation rearranges to \[(x^3+1)[P(x+1)-2P(x)+P(x-1)] = mx[x(P(x+1)-P(x-1))-2P(x)].\]Replacing finite differences with regular derivatives, we see that the right side (without the $mx$) becomes \[2xP'(x)-2P(x),\]which can be written as $2x^2(P/x)'$. This motivates looking at $Q(x)$ to be the finite difference of $P(x)/x$, so \[Q(x):=\frac{1}{x+1}P(x+1)-\frac{1}{x}P(x).\]Indeed, it is not hard to check that given this substitution, we have that the left hand side is \[m(x^3-x)[Q(x)+Q(x-1)],\]which matches our $2mx^3(P/x)'$ prediction. The next part is to right the right hand side in terms of $Q$, which in the derivative analogy amounts to writing $P''$ in terms of $Q$. Indeed, the simplest way to do this is \[P''=(xQ)'+Q.\]Given this, it is not too hard to get that \[P(x+1)-2P(x)+P(x-1) = [xQ(x) - (x-1)Q(x-1)] + Q(x),\]which again matches the above prediction. The beauty of this setup is that only $Q(x)$ and $Q(x-1)$ are involved, so we can directly solve for $Q(x)/Q(x-1)$ as a rational function. The introduction of $R(x)$ is now just to do this more cleanly from the start already knowing the result through our $Q$ analysis.
cefer
25.04.2021 21:43
Here is a totally different approach. It is easy to check that if $n = \deg P \leq 1$, then $P(x) = cx$ for some constant $c$. If $n\geq 2$, then by comparing coefficients of $x^{n+1}$ on both sides of the given equation we get $n = 2m$. Now, replace $x$ with $-x$ \[ (x^3 + mx^2 - 1 ) P(-x+1) + (x^3-mx^2-1) P(-x-1) =2(x^3 - mx -1 ) P(-x) \]and add this to the given equation. On the left, the first summand in the main equation and the second summand in above equation can be combined to \[ (x^3 - mx^2 )\left(P(x + 1) + P(-x-1)\right) + P(x + 1) - P(-x-1)\]remaining terms on the left side are \[(x^3+mx^2) \left(P(x-1) + P(-x+1)\right) + P(x-1) - P(-x+1)\]and on the right hand side we have \[2(x^3 - mx )\left(P(x) + P(-x)\right) + P(x) - P(-x)\]Note that $Q(x) + Q(-x)$ is even and $Q(x) - Q(-x)$ is odd polynomial for any polynomial $Q$. So while on the right hand side we only have odd polynomials (see the third group above), on the left only products with $mx^2$ yield even terms (see the first and second groups above). Since $m\neq 0$ we get \[-P(x + 1) - P(-x-1) + P(x-1) + P(-x+1) = 0\]Thus \[P(x-1) + P(-x+1) = P(x + 1) + P(-x-1)\]So, if $R(x) = P(x-1) + P(-x + 1)$, then $R(x) = R(x+2)$ and since $n$ is even we have $\deg R = n$. But this is a contradiction because $R$ is a nonzero polynomial with infinite roots.
alexiaslexia
07.06.2021 16:33
The solutions to the equation are $\boxed{P(x) = cx}$ for every $c \in \mathbb{R}$. It is easily verified that this satisfies the problem statement. Did not expect this to be a monstrous problem. Here's a Solution without the magical
$xP(x+1) - (x+1)P(x)$
substitution (this does capitalise on that substitution, implicitly.) To establish that all polynomials that satisfy the equation is $\color{green} \textbf{\textit{linear with a zero constant}}$, we prove the following Claim. $\rule{5.15cm}{2pt}$ $\diamondsuit$ $\boxed{\textbf{The Elusive End Goal.}}$ $\diamondsuit$ $\rule{5.15cm}{2pt}$ We prove that there exists a number $l \in \mathbb{C}$ so that \[ P(x) - lx \]has infinitely many zeroes.
If this is true, then $P(x) \equiv lx$; as the polynomial $P(x)-lx$ with infinite (not necessarily real) distinct roots must be the zero polynomial.
Since $P(x) \in \mathbb{R}[x]$, $l$ must be real to satisfy the problem statement.
$\rule{25cm}{0.2pt}$ Indeed, we Claim that there exists a $r \in \mathbb{C}$ so that either \[ r-1,r,r+1,\ldots,r+n ,\ldots \]or \[r+1,r,r-1,\ldots,r-n,\ldots \]are all zeroes of $P(x) - lx$.
This should give an insight on why the problem gives only $P(x+1),P(x),P(x-1)$ and why $m$ should be an integer --- we exclusively rely on $P$'s recursive properties.
To put it simply, this is actually a sequence problem with complex roots (and explicit division of terms).
To start, we manipulate the original assertion to make it more accessible. $\color{green} \rule{5cm}{2pt}$ $\color{green} \diamondsuit$ $\boxed{\textbf{Crucial Manipulation.}}$ $\color{green} \diamondsuit$ $\color{green} \rule{5cm}{2pt}$ Separate the $x^3+1$ and $mx$, resulting in \[ (x^3+1) (P(x+1)+P(x-1)-2P(x)) = mx (xP(x+1)-xP(x-1)-2P(x)). \]Since $\gcd{(x,x^3+1)} = 1$, then \[ x \mid P(x+1)+P(x-1)-2P(x). \]Writing $P(x+1)+P(x-1)-2P(x) = mx \cdot Q(x)$, we have \[ xP(x+1)-xP(x-1)-2P(x) = (x^3+1) Q(x). \]We rely on these two equations to find the beautiful number $r$. For now, we are content with these friendlier expressions. $\blacksquare$ $\color{magenta} \rule{7.7cm}{2pt}$ $\color{yellow} \diamondsuit$ $\boxed{\textbf{Performing Miracles with the Roots.}}$ $\color{yellow} \diamondsuit$ $\color{magenta} \rule{7.7cm}{2pt}$ Let $Q(x)$ be a nonconstant polynomial.
Then, it is easy to see that $P$'s degree is at most $3$ (since $P(x+1)-2P(x)+P(x-1)$ has two less degree than $P$, and it has degree $1$, since $mx \cdot Q(x)$ is a linear polynomial.
We can just bash this.
Then, $Q$ must have roots (a very loose assumption, is it not?) Let one of them be $r$. $\color{magenta} \rule{25cm}{0.2pt}$ Now we know that $P(r+1)+P(r-1)-2P(r)$ and $rP(r+1)-rP(r-1)-2P(r)$ both equals zero. As a result, we have \[ P(r+1)+P(r-1) = rP(r+1)-rP(r-1) \]or \[ (r-1)P(r+1) = (r+1)P(r-1). \]If $P(r+1) = P(r-1)=0$, set $l = 0$. Using $P(r+1)+P(r-1)-2P(r) = 0$ we know that $P(r-1) = P(r) = P(r+1) = 0$. If both $P(r+1),P(r-1)$ are nonzero, we know that \[ \dfrac{P(r+1)}{r+1} = \dfrac{P(r-1)}{r-1} = L. \]Set $l = L$, and we can assume Without Loss of Generality that $P(x)-lx$ is the new $P(x)$. (If you are bothered by this, then let this be $P_1$ and replace all $P$ below by $P_1$; those statements are true since $P_1$ also satisfies the original assertion as $P$ does.) Since this implies $P(r-1) = P(r) = P(r+1) = 0$ too, we are done establishing the base case towards $\diamondsuit$ $\boxed{\textbf{The Elusive End Goal}}$ $\diamondsuit$'s conclusion. $\color{magenta} \rule{25cm}{0.2pt}$ But wait! What happens if $r-1$ or $r+1$ equals zero?
If $r = 1$ or $r = -1$, subbing this to
\[ (r-1)P(r+1) = (r+1)P(r-1) \]yields $P(0) = 0$. One could just bash as in $\#6$, letting $P(x) = xR(x)$. However $P(0) = 0$ yields an easy recursion: let $l = \frac{P(1)}{1}$ (or just plain zero if $P(1)$ is already zero from the start) and we would have
\[ P(0) = P(1) = 0 \]After this, we can simply prove that $P(n) = 0$; as we know that
\[ x^3-mx^2+1 \ne 0 \ \text{for} \ x \geq 1\]except for $m = 2$!!! (Annoying, right?)
In that case, just let $l = \frac{P(-1)}{-1}$ (same convention if $P(-1) = 0$ as above) and just prove that $P(-n) = 0$.
We are done for the second part. $\blacksquare$ $\blacksquare$
While this Claim is certainly difficult to find (especially since the expressions aren't as symmetric and managable as we would like), I did get stuck on this for a long time.
The trickiest (and easily fakesolve-inducing) part was the inclusion of $r$ inside the nice-root-set; if we exclusively rely on the original assertion, $r$ can actually be a root of $x^3-mx+1$. If this happens, then while the LHS is certainly zero if
\[ P(r-1) = P(r+1) = 0 \]there is no way to control $P(r)$'s value if $r^3-mr+1$ is zero.
On the other hand, if we rely on
\[ P(x+1)+P(x-1)-2P(x) \]being zero, we would be set with no roadblock; the remaining journey would only consist of picking one direction to traverse into.
Time for the heavy computation! $\color{blue} \rule{4.9cm}{2pt}$ $\color{blue} \clubsuit$ $\color{blue} \boxed{\textbf{Expanding to Infinity.}}$ $\color{blue} \clubsuit$ $\color{blue} \rule{4.9cm}{2pt}$ We Claim that $r+a$ and $r+b$ for some $a,-b \in \mathbb{N}$ cannot be a root of $x^3-mx^2+1$ and $x^3+mx^2+1$, respectively. $\color{blue} \rule{4.9cm}{0.2pt}$ If this is so, then the sequence cannot be a dead end in both directions. So, if no number in the form of $r+a$ is a root of the first polynomial, we would have $P(r+a) = 0$ given that \[ P(r+a-1), P(r+a-2) = 0. \]Since we already have $P(r),P(r+1) = 0$, we are set to venture towards $r+ (+\infty)$ by simple recursion. Likewise for $P(r+b)$ --- we can have $P(r+b) = 0$ since we already have $P(r),P(r-1) = 0$. $\color{blue} \rule{25cm}{0.2pt}$ $\color{blue} \spadesuit$ $\color{blue} \boxed{\textbf{Proof.}}$ $\color{blue} \spadesuit$ (Almost) mindless bashing. Let we have \[ x^3-mx^2+1 = 0, \ \text{and} \ y^3+my^2+1 = 0\]with $x-y = z \in \mathbb{Z}$.
Substituting $x = y+z$ gives
\[ y^3+my^2+1-2my^2+(3y^2z+3yz^2+z^3)-(2myz+2mz^2) = 0. \]This implies
\[ (3z-2m)y^2+(3z^2-2mz)y+(z^3-mz) = 0. \]Multiplying this by $y$ and substracting it with $y^3+my^2+1 = 0$ multiplied by $(3z-2m)$, we get
\[ (3z^2-5mz+2m^2)y^2+(z^3-mz^2)y-(3z-2m) = 0. \]Substracting this with the penultimate equation multiplied by $(z-m)$, we get
\[ (\text{something} \cdot y) - (z^2(z-m)(z-2m)+(3z-2m)) = 0. \]If the last term is nonzero, we get $y \in \mathbb{Q}$ or $y$ doesn't exist.
Last Term Zero.We get that
\[ z^2 \mid 3z-2m, z-m \mid m. \]From the second assertion, we have $m = k \cdot (z-m)$, and $m = \frac{k}{k+1}z$. Substituting that to the first assertion, we get
\[ z \mid \frac{k+3}{k+1}. \]As $k+1 \mid 2$, $k \in \{-2,-1,1,2\}$. This implies $z \mid -1,0,2,3$. Solving all six cases for $m$ yields no solutions; $z = 0$ implies a contradiction since $m \ne 0$.
If $y \in \mathbb{Q}$, $x$ is also rational. This implies $\{x,y\} = \{1,-1\}$, and $r \in \mathbb{Z}$. To get a full clear, proceed as in the earlier $m = -2,2$ case (in $\text{hide = Annoying computation}$. )
We are finally done. $\blacksquare$ $\blacksquare$ $\blacksquare$
Basically, I just had to find the manipulation/algebraic idea which narrowed down two consecutive values of $P(x)$. Say if I had $P(x) = Cx$ and $P(x+1) = C(x+1)$ (consecutive means numbers with difference $1$), there's a big chance that picking $l = C$ would generate infinite roots of $P(x)$.
$\color{green} \rule{25cm}{2pt}$
$\textbf{\textit{Before Simplification}}$
$\color{red} \text{Sec 1: Shifting Brothers.}$ Immediately, after knowing that $P(x) = cx$ works for every $c \in \mathbb{R}$, and $P(x) = cx+d$ $\textbf{\textit{could not work}}$ for any $d \in \mathbb{R}$, I knew that the usual methods will not work. (The set of usual methods consists of regular degree subbing, bashing coefficient/roots, and arbitrarily deducing from some special property.)
In contrast, the problem already provides a seemingly unrelated info that $m \in \mathbb{Z}$; and you don't normally see extraneous info on an $\text{A6}$ For comparison, an example of a problem with extraneous info is
SPOILER for a famous international problem.IMO 2020 P4.
Proceeding from that, I judged that the best way to proceed is by letting
\[ Q_1(x) P(x+1) + Q_2(x) P(x-1) = Q_3(x) P(x) \]and forcing some constraint on $Q_i$'s existence. However, $\color{red} \text{the one}$ (obvious) singular expression
\[ Q_1 \cdot (x+1) + Q_2 \cdot (x-1) = Q_3 \cdot x \]can only eliminate one degree of freedom, and $P$ supposedly has $d$ degrees of freedom ($d$ being its degree).
Next, as with any polynomial problem with a $\textit{simple}$ condition, I tried to bash out possible expansions. This does nothing good on $P \equiv ax^2$ and later (just try plugging that towards the original equation, and you get a mess of numbers added with $m$.)
My final attempt before stumbling on the (almost) correct one was explicitly narrowing our attention to $a_d$ ($P$'s leftmost coefficient) and its binomial madness; but I found nothing good on the expression
\[ Q_1 \cdot (x+1) + Q_2 \cdot (-x+1) \]since we would have to rely on $Q_2$ multiplied by $-x+1$ (not only by $x-1$). Here I was again faced with the problem of $\color{red} \text{missing degree of freedom.}$
Seeing that $P \equiv ax^2+bx+c$ already creates a wild ride if I were to expand it, I opted to exclusively focus on $Q_i$. Luckily, this put the weight on $x^3-mx^2+1$, $x^3+mx^2+1$ and $x^3-mx+1$'s standings with one another.
Consequently, I started to separate these convoluted additions into $A(x)$ and $B(x)$ where
\[ Q_1 = A-B, Q_2 = A+B, Q_3 = A - \frac{B}{x} \]where $A = x^3+1$ and $B = mx^2$. Grouping like terms together (I finally am successful in grouping $\textbf{some}$ terms of the expression), the expression becomes
\[ A(x) [P(x+1)+P(x-1)-2P(x)] = B(x) \left[ P(x+1)-P(x-1)-\dfrac{P(x)}{x} \right] \]This is especially nice since it preserves the structure which makes $cx$ a viable solution (while $cx+d$ is not) without losing any important info about $A$ and $B$ were we are required to analyze their property later.
Notes 4.The structure is what I called $\textbf{the shifting brothers}$ as in the title; upon their respective multiplications with $x+1,x-1$ and $x$ respectively, the term $x^3+1$ successfully moved one space forwards and $mx^2$ stayed in place.
The final manipulation after this brings them together again by regressing $mx^2$ to fit with $x^2+1$'s pace by lowering it (by dividing with $x$) into $mx$.
To simplify, since $B$ has an $x^2$ and the terms between the square brackets have a denominator of $x$, I simplified them to the form I finally use in the Solution.
$\color{green} \rule{25cm}{2pt}$
$\textbf{\textit{After Simplification}}$
$\color{red} \text{Sec 2: Unfrindly Expression? Zero It!!}$ After explicitly using $Q(x)$ as the Solution does, I immediately tried to plug $r$ where $Q(r) = 0$ for dramatic effect.
This yields
\begin{align*}
P(r+1) + P(r-1) &= 2P(r) \\
rP(r+1) - r P(r-1) &= 2P(r)
\end{align*}this was quite uninteresting for my eyes, so I did what the sixth-grader would do: since $Q(x)$ was supposed to be zero (try to plug $P \equiv cx$ into the new form) why not just consider the degree of $Q$? Here, I saw $\textit{degree}$ as the number of roots of a polynomial; if $Q$'s destiny is to be zero, won't simple coefficient bash solve the problem?
Turns out that again, even after using this form (even after using the forms where they only utilize $Q(x)$ and $Q(x-1)$) there's no easy way to deduce something only from looking at the coefficients' magnitude.
Here, the most notable observation that all terms from both sides (excluding from $mx$ and $x^3+1$) was that all coefficients are of the same parity-degree. Moreover, they have equal sums, except for $\color{red} \text{the pesky constant}$. When comparing, subbing $P \equiv 1$ yields a zero on the LHS (a.k.a. the expression $P(x+1)+P(x-1)-2P(x)$) and a $-2$ on the RHS.
Whenever I tried to equate the even and odd coefficients, they are supposed to be equal (and force $m = 2$ or $m = -2$), save for the constant's intervention. (And the approach to consider the parity of $P$'s coefficients $\textbf{fails spectacularly}$, because they are $\textbf{real numbers}$, not integers.)
With a renewed vigor to prove $P(0) = 0$, I quickly noticed that this alone would solve the problem. If it is so, we can supposedly set $P(x) - P(1)x$ to be the new $P$ and proceed directly to $\diamondsuit$ $\boxed{\textbf{The Elusive End Goal}}$ $\diamondsuit$ by recursion.
Notes 5.This is how this Section got its name. If only we could find $P(r)$ and $P(r+1)$ with equal gradient w.r.t. the origin, we can easily induct up or down!
The problem towards this is always the same; we are constantly teased that if two consecutive terms behave nicely, then we are mostly done if $\color{blue} \clubsuit$ $\color{blue} \boxed{\textbf{Expanding to Infinity}}$ $\color{blue} \clubsuit$ happens smoothly.
However, there is a side of me that rejects the possibility of $P(x) \equiv cx$ to be the only solution: what if the polynomial actually has two degrees of freedom and we just have to adjust $P(0)$ and $P(1)$ to them to necessarily facilitate the safe expand?
After bashing the values of $P(-2)$ until $P(0)$ using only the two forms in $\color{green} \diamondsuit$ $\boxed{\textbf{Crucial Manipulation}}$ $\color{green} \diamondsuit$ (i.e. if $P(1) = 0$, then $P(-0) = c$, $P(-1) = 2c$ and $P(-2) = 5c$) I gave up entirely on controlling only the images of integers under $P$.
Instead, I noted that $r$ could very well be any real number. I considered the entire number line for $r$ where
\[ \dfrac{P(r)}{r} = \dfrac{P(r+1)}{r+1} \]but some polynomials such as $P(x) = x^3$ just don't have two easy differences. (I only learnt later that taking a random root of $xP(x+1) - x+1 P(x)$ works; and $P(x) = x^3$ does indeed have an easy $r$.)
Since I was desperate, I scanned the entire complex plane for a possible $r$ --- and found that
\[ (r-1) P(r+1) = (r+1) P(r-1) \]and thought that it was good enough. Coincidentally, the two numbers with $2$ spaces separating it can be condensed into three numbers with exactly $1$ space between them by adding $P(r) = 0$ into the mix!
From here on, it's just <15 MOHS bash. I really liked the idea of using rational root theorem, $Q_1,Q_2,Q_3$ not being possible to explicitly compute, and learning to juggle polynomials into their most optimal form.
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vsamc
26.07.2024 22:36
The answer is $\boxed{P(x) = cx}$, where $c\in \mathbb{R}$. This can easily be checked to work. We now show that these are the only such polynomials that work. First of all, note that the only polynomials with degree $\leq 1$ that work are the one of the form $P(x)=cx$, since if $P(x)=ax+b$ works, then since $ax$ works, it's clear that $P(x)-ax = b$ works (since, the set of all polynomials that work is clearly a vector space), but evidently no nonzero constant solutions work because if $P\equiv c$ the LHS is $c(2x^3+2)$ and the RHS is $c(2x^3-2mx+2)$. Now, assume for the sake of contradiction that there existed a polynomial $P$ that worked that had degree $\geq 2$. First, we claim the following:
There does not exist $r, s$ such that $r-s\in \mathbb{Z}$ and $r^3-mr^2+1=s^3+ms^2+1=0$. To prove this, first note that $r$ and $s$ are algebraic numbers with minimal polynomials in $\mathbb{Q}[x]$ having degree at most $3$. In fact, since $s = r+n$ for some integer $n$, we have that $$[r^3-mr^2+1] - [(r-n)^3+m(r-n)^2+1] = 0,$$which is at most quadratic in $r$, so the minimal polynomial of $r$ in $\mathbb{Q}[x]$ must have degree at most $2$, and since $s=r+n$ for $n\in \mathbb{Z}$, it follows that the minimal polynomial of $s$ in $\mathbb{Q}[x]$ must have degree at most $2$ as well. If the minimal polynomial of $r$ in $\mathbb{Q}[x]$ has degree $1$, then the same must be true for $s$, and this implies that $r$ and $s$ are rational. By RRT this implies $r,s\in \{-1, 1\}$. We cannot have $r$ or $s$ be equal to $-1$ because it would imply $m=0$. So $r=s=1$. However, $r=1$ implies $m=2$, but $s = 1$ implies $m=-2$, absurd. So, the minimal polynomial of $r$ in $\mathbb{Q}[x]$ has degree $2$, and the same must hold for $s$. Then, the minimal polynomial of $r$ in $\mathbb{Q}[x]$ must divide $x^3-mx^2+1$, and the quotient will be a linear polunomial in $\mathbb{Q}[x]$, so $x^3-mx^2+1$ has a rational root. By similar logic with $s$ instead of $r$, $x^3+mx^2+1$ must also have a rational root. But we just showed that both of these happening simultaneously was impossible in the above case. So, we have a contradiction, and so the claim is true.
Now, note that the given identity holding for all $x\in \mathbb{R}$ implies that it holds for all $x\in \mathbb{C}$ because it is a polynomial identity (the difference is a polynomial with infinitely many roots, so it is identically $0$ on any infinite subset of $\mathbb{C}$). Now, let $Q(x) = (x+1)P(x)-xP(x+1)$. Let $P(x) = a_d x^d + a_{d-1}x^{d-1} + \cdots + a_1 x + a_0$, for some $d\geq 2$, where $a_d\neq 0$. Then, note that the $x^{d+1}$ term cancels out in $Q$, so the leading term of $Q$ is $$(a_{d-1}+a_d)x^d - (da_d+a_{d-1})x^d = a_d(1-d)x^d,$$and since $d\geq 2$ and $a_d \neq 0$, this term is nonzero, so $\deg Q \geq d$. Thus, since $d\geq 2$, it follows that $Q$ is nonconstant, so it has a root $z\in \mathbb{C}$. Then, we have that $$0 = Q(z) = \det \begin{vmatrix} P(z) & P(z+1) \\ z & z+1 \end{vmatrix},$$so $\langle P(z), P(z+1) \rangle$ and $\langle z, z+1\rangle$ are linearly dependent, so either one of them is the $0$ vector or they are multiples of each other. Clearly $\langle z, z+1\rangle$ is not the zero vector, so either $\langle P(z), P(z+1)\rangle = c\langle z, z+1\rangle$ for some $c\in \mathbb{C}$ or $\langle P(z), P(z+1)\rangle$ is the $0$ vector, but if we have the latter case, we can select $c=0$. So, there exists a $c\in \mathbb{C}$ such that $\langle P(z), P(z+1)\rangle = c\langle z, z+1\rangle$, i.e. $P(z) = cz$ and $P(z+1) = c(z+1)$. Now, note that if $a(x)=x^3-mx^2+1$ and $b(x) = x^3+mx^2+1$, then at least one of $\{a(z+1), a(z+2), a(z+3), \cdots \}$ and $\{b(z), b(z-1), b(z-2), \cdots \}$ is $0$-free (i.e. it does not have $0$ as an element), because if there existed an integer $r\geq 1$ such that $a(z+r) = 0$ and an integer $s\geq 0$ such that $b(z-s) = 0$, then we violate the above claim since $(z+r)-(z-s) = r+s\in \mathbb{Z}$. Now, we split into cases.
Case 1: $\{a(z+1), a(z+2), a(z+3), \cdots \}$ is $0$-free.
Then, we will prove by (strong) induction that $P(z+n) = c(z+n)$ for all integers $n\geq 0$. Our base cases are $n=0, 1$, which we've already established. Now, assume that for some $k\geq 1$, we have that $P(z+j) = c(z+j)$ for all $0\leq j\leq k$. Then, if we let $z' = z+k$, we have that $P(z') = cz'$ and $P(z'-1) = c(z'-1)$ by the inductive hypothesis, so plugging in $x = z'$ (remember, we showed the condition holds for all $x\in \mathbb{C}$),
\begin{align*}
(z'^3-mz'^2+1)P(z'+1) + (z'^3+mz'^2+1)P(z'-1) &= 2(z'^3-mz'+1)P(z') \\
\implies (z'^3-mz'^2+1)P(z'+1) + (z'^3+mz'^2+1)(c(z'-1)) &= 2(z'^3-mz'+1)(cz') \\
\implies (z'^3-mz'^2+1)P(z'+1) &= (z'^3-mz'^2+1)(c(z'+1)) \\
\implies P(z'+1) &= c(z'+1),
\end{align*}where we can safely divide by $z'^3-mz'^2+1$ on the last step because $z'^3-mz'^2+1 = a(z+k) \neq 0$, as $\{a(z+1), a(z+2), a(z+3), \cdots \}$ is $0$-free. So, $P(z'+1) = c(z'+1)$, i.e. $P(z+k+1) = c(z+k+1)$. Thus, the induction is complete and so $P(z+n) = c(z+n)$ for all integers $n\geq 0$. Thus, $P(x)-cx$ has infinitely many roots, so $P(x) = cx$ for all $x\in \mathbb{C}$. However, since we are assuming $\deg P \geq 2$, this is a contradiction.
Case 2: $\{b(z), b(z-1), b(z-2), \cdots \}$ is $0$-free.
Then, we will prove by (strong) induction that $P(z-n) = c(z-n)$ for all integers $n\geq -1$. Our base cases are $n = -1, 0$, which we've already established. Now, assume that for some $k\geq 0$, we have that $P(z-j) = c(z-j)$ for all $-1\leq j\leq k$. Then, if we let $z' = z-k$, we have that $P(z') = cz'$ and $P(z'+1) = c(z'+1)$ by the inductive hypothesis, so plugging in $x=z'$,
\begin{align*}
(z'^3-mz'^2+1)P(z'+1) + (z'^3+mz'^2+1)P(z'-1) &= 2(z'^3-mz'+1)P(z') \\
\implies (z'^3-mz'^2+1)(c(z'+1)) + (z'^3+mz'^2+1)P(z'-1) &= 2(z'^3-mz'+1)(cz') \\
\implies (z'^3+mz'^2+1)P(z'-1) &= (z'^3+mz'^2+1)(c(z'-1)) \\
\implies P(z'-1) &= c(z'-1),
\end{align*}where we can safely divide by $z'^3+mz'^2+1$ on the last step because $z'^3+mz'^2+1 = b(z-k) \neq 0$, as $\{b(z), b(z-1), b(z-2), \cdots \}$ is $0$-free. So, $P(z'-1) = c(z'-1)$, i.e. $P(z-k-1) = c(z-k-1)$. Thus, the induction is complete and so $P(z-n) = c(z-n)$ for all integers $n\geq -1$. Thus, $P(x)-cx$ has infinitely many roots, so $P(x)=cx$ for all $x\in \mathbb{C}$. However, since we are assuming $\deg P \geq 2$, this is a contradiction.
Thus, in all cases, we have a contradiction when $\deg P \geq 2$, so the only solutions are the ones with $\deg P \leq 1$, which we've already found, so we're done. $\blacksquare$
awesomehuman
24.08.2024 06:53
Rearrange the equation to get \[(x^3+1)(P(x+1)+P(x-1)-2P(x)) = mx(xP(x+1)-xP(x-1)-2P(x)).\]Thus, for some polynomial $R(x)$, \[P(x+1)+P(x-1)-2P(x) = 2mxR(x)\]\[xP(x+1)-xP(x-1)-2P(x) = 2(x^3+1)R(x).\] Therefore, \[xP(x+1)-(x+1)P(x) = (x^3+mx^2+1)R(x)\]\[(x-1)P(x)-xP(x-1) = (x^3-mx^2+1)R(x).\] So, \[(x^3+mx^2+1)R(x) = ((x+1)^3-m(x+1)^2+1)R(x+1).\]Let $Q(x)= xP(x+1) - (x+1)P(x)$. Assume toward a contradiction $R\neq 0$. Let $r$ be a root of $R(x)$. Let $r_1$ and $r_2$ be the minimum and maximum values, respectively, such that $r_1$ and $r_2$ are roots of $R$ and $r_1-r$ and $r_2-r$ are integers. Then, \[((r_2+1)^3-m(r_2+1)^2+1)R(r_2+1) = (r_2^3+mr_2^2+1)R(r_2) = 0.\]Therefore, $((r_2+1)^3-m(r_2+1)^2+1) = 0$. We also have \[((r_1-1)^3+m(r_1-1)^2+1)R(r_1-1) = (r_1^3-mr_1^2+1)R(r_1) = 0.\]Therefore, \[((r_1-1)^3+m(r_1-1)^2+1) = 0.\]Let $a = r_1-1$ and $k = r_2 + 1 - a$. Then, $a$ is a root of $x^3+mx^2+1$ and $a+k$ is a root of $x^3-mx^2+1$. So, we have \[(a+k)^3 - m(a+k)^2 + 1 = 0\]\[a^3-ma^2+1 +3ka^2+3k^2a+k^3 -2mka+mk^2 = 0\]\[-2ma^2+3ka^2+3k^2a+k^3 -2mka+mk^2 = 0\] Therefore, $a$ is a root of a degree $2$ integer polynomial. So, $x^3-mx^2+1$ is not irreducible. It is well known that if an integer polynomial factors into rational polynomials, it factors into integer polynomials. So, $x^3-mx^2+1=(x-t)f(x)$ for some integer $t$ and quadratic $f(x)$. Therefore, $t^3-mt^2+1=0$. So, $t=\pm 1$. Therefore, $m=2$ and $t=1$. Because $a$ is a root of a degree $2$ integer polynomial, $a+k$ also is. So, $x^3+mx^2+1$ is also not irreducible. By similar logic to before, $m=-2$, a contradiction. Therefore, $R(x)=0$. So, \[xP(x+1)=(x+1)P(x)\]\[\frac{P(x+1)}{x+1} = \frac{P(x)}{x}.\]So, for some $c$, $\frac{P(n)}n=c$ for all positive integers $n$. Thus, $P(n)=nc$ has infinitely many roots, so $P(x)=cx$. Clearly this works.