Determine all polynomials $P(x)$ with real coefficients such that $(x+1)P(x-1)-(x-1)P(x)$ is a constant polynomial.

  • clearly we have to show $(x+1)P(x-1)-(x-1)P(x)=c$ for all values of $x$ ($c$ is a real constant)
  • i have tried this: see if $r$ is a root of $P(x)$ then $P(r)=0$ and then $(r+1)P(r-1)=c$. Now we consider $P(x)=a(x-r)(x-c)(x-d)\cdots(x-l)$, so we have $$(r+1)P(r-1)=(c+1)P(c-1)=(d+1)P(d-1)=\dots\\=(l+1)P(l-1)=\pm 2rcd\dots l=c$$ [because see $P(0)=P(-1)=\dfrac{c}{2}$ ]
  • then i am stuck!!!!!!! see $P(x)=v$ is a solution where $v$ is a constant.
  • $\begingroup$ This condition will translate into linear equations on the coefficients. It is then a matter of checking if that linear system is consistent. $\endgroup$ Mar 26, 2014 at 8:20
  • $\begingroup$ i couldn't understand.please answer mathematically. $\endgroup$ Mar 26, 2014 at 12:23
  • $\begingroup$ Since $c$ will vary with $P$, it's probably easier to solve $G'(x)=0$, where $G(x)=(x+1)P(x-1)-(x-1)P(x)$. $\endgroup$
    – lhf
    Mar 26, 2014 at 12:39

2 Answers 2


As Vanchinathan said in the comments,

Take $P(x)= \sum\limits_{n=0}^r a_n x^n$ then plug in this into the equation

you get $\sum\limits_{n=0}^r a_n [ (x+1)(x-1)^n-(x-1)x^n]=c$

so $2 a_0 + a_1 (x-1) + \cdots+ a_r [ (x+1)(x-1)^r-(x-1)x^r]=c$ and

then you create linear equations with the coefficients using polynomial equality with the RHS. I hope this helps a bit.


Let $(x+1)P(x−1)−(x−1)P(x)=C=const$. Set $x=2$ in order to get $$P(2)=3P(1)-C.$$ Set $x=3$ to obtain $$P(3)=6P(1)-\frac{5}{2}C.$$ This motivates us to try to prove by induction that $$P(n)=\frac{n(n+1)}{2}P(1)-\frac{(n-1)(n+2)}{4}C$$ for every integer $n\geq 2.$ Indeed, if we assume the last equality holds for some $n\geq 2,$ then for $n+1$ we just set $x=n+1$ and obtain $$P(n+1)=\frac{1}{n}\bigg( \frac{n(n+1)(n+2)}{2}P(1)-\frac{(n-1)(n+2)^2+4}{4}C \bigg)=\frac{(n+1)(n+2)}{2}P(1)-\frac{n(n+3)}{4}C$$ as desired. But since $P$ agrees with a quadratic polynomial at infinitely many integer values, we conclude that $P$ is quadratic. If we then let $P(x)=ax^2+bx+c,$ we can plug it into the original equation and see that all solutions are of the form $$P(x)=ax^2+ax+\frac{C}{2},$$ where $a$ is an arbitrary real number and $C$ is the constant given in the beginning.


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