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Find all polynomials satisfying the functional equation $$(x+1)P(x) = (x-10)P(x+1).$$

I'm studying polynomials from the book Putnam and Beyond and for this particular question they had something which I've have never seen before. The solution was based on something called "Shifting the variable" where they got from the original equation to $xP(x-1)=(x-11)P(x).$ Now I don't now what kind of magic is this so could someone enlighten me on what is happening here and if there would be some other way to approach the problem, that would be highly appreciated.

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    The "shifting the variable" done to get the equation $xP(x-1)=(x-11)P(x)$ is "replacing $x$ by $x-1$". The original equation is valid for all $x$, so it must also be valid for all $x-1$, therefore this step is allowed. – Wolfgang Kais Jun 23 '20 at 23:36

4 Answers4

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In this case, they have shifted the variable by $1$, or changed $x\iff x+1$

It might be easier to see if you impose $y=x+1$ into the original equation, so you get $$yP(y-1)=(y-11)P(y)$$ which matches their conclusion but with $y$ instead of $x$.

This is a legitimate step in a functional equation like this, because the functions are defined everywhere, so if it works for $x$, it too works for $x+1$.

Rhys Hughes
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hint

If $ x=-1 $, it gives $$0=-11P(0) \implies P(0)=0$$

If $ x=0 $, we get $$P(0)=-10P(1)=0 \implies P(1)=0$$

hamam_Abdallah
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$(x-10)$ is a factor of $P(x)$. Now sub this into the equation ... so

$(x-9)$ is a factor of $P(x)$. Now sub this into the equation ... so

...

And finally $P(x)=A(x-10)(x-9) \cdots (x-1)x$.

Donald Splutterwit
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Hint.

Making $P(x) = \sum_{k=0}^n a_k x^k$ and substituting into the relationship we conclude that $n=11$ and

$$ P(x) = C_0\left(\frac{x^{11}}{3628800}-\frac{11 x^{10}}{725760}+\frac{11 x^9}{30240}-\frac{121 x^8}{24192}+\frac{7513 x^7}{172800}-\frac{8591 x^6}{34560}+\frac{341693 x^5}{362880}-\frac{84095 x^4}{36288}+\frac{177133 x^3}{50400}-\frac{7381 x^2}{2520}+x\right) $$

Calling $A = \left(a_0,\cdots,a_{11}\right)$ and

$$ M = \left( \begin{array}{cccccccccccc} 11 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 & 10 \\ 0 & 10 & 19 & 29 & 39 & 49 & 59 & 69 & 79 & 89 & 99 & 109 \\ 0 & 0 & 9 & 27 & 56 & 95 & 144 & 203 & 272 & 351 & 440 & 539 \\ 0 & 0 & 0 & 8 & 34 & 90 & 185 & 329 & 532 & 804 & 1155 & 1595 \\ 0 & 0 & 0 & 0 & 7 & 40 & 130 & 315 & 644 & 1176 & 1980 & 3135 \\ 0 & 0 & 0 & 0 & 0 & 6 & 45 & 175 & 490 & 1134 & 2310 & 4290 \\ 0 & 0 & 0 & 0 & 0 & 0 & 5 & 49 & 224 & 714 & 1848 & 4158 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 4 & 52 & 276 & 990 & 2838 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 3 & 54 & 330 & 1320 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 55 & 385 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 55 \\ \end{array} \right) $$

we have $M\cdot A = 0$

Cesareo
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