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Similar to Find polynomials : $ xP(x-1)=(x-11)P(x)$ but my solution is different.

Bob asks us to find all real polynomials $P$ such that $$xP(x-1)=(x-2)P(x)$$

Here is my solution, is it right?

Replace $x$ with $0$ and $1$ and clearly, they equal $0$.

Suppose there is another positive root $r\neq 1$.

Then replace $x$ with $r+1$, which means $(r+1)P(r)=(r-1)P(r+1)\implies P(r+1)=0$ since $r-1$ and $r+1$ are not $0$.

Thus if there exists another positive root $r\neq 1$, then we have an infinite number of roots which means $P(x)=0$.

Now suppose there is another negative root $c$.

Replace $x$ with $c$, sos $cP(c-1)=(c-2)P(c)\implies P(c-1)=0$ since $c$ and $c-2$ are not $0$.

Thus if $c$ exists, then we have an infinite number of roots which again means $P(x)=0$.

So the only two polynomials that work are $P(x)=x(x-1)$ and $P(x)=0$.

nonuser
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user627514
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  • Why is $P(r+1)=0$? – nonuser Dec 31 '18 at 12:36
  • It is better if you go by carat foot steps... – nonuser Dec 31 '18 at 12:37
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    What you showed is the following: either $P(x)$ is the constant zero polynomial, or it has no integer root different from $0$ and $1$. However, you cannot conclude the result you mentioned from this observation. Based on this observation, you can only conclude that $P(x)=x(x-1)Q(x)$, such that $Q(x)$ has no integer root. There are many such $Q(x)$, how would you know that only $Q(x)=1$ is correct. By the way, it isn't: $Q(x)$ can be any nonzero constant, for example. – A. Pongrácz Dec 31 '18 at 12:59

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If you put $x=2$ you get $P(1)=0$ and if you put $x=0$ you get $P(0)=0$ so 1 and 0 are zeroes so we can write: $$P(x)= x(x-1)Q(x)$$ for some polynomial $Q$. If you put this in original equation we get $$Q(x)=Q(x-1)$$ which is valid for all $x\ne 0,1,2$ and there for it is valid for all $x$, so $Q$ must be constant (since it is periodic).

So general soution is $$P(x) = cx(x-1)$$ for arbitrary constant $c$.

nonuser
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  • Was about to upvote when I realsied that you're not actually answering the question. – Git Gud Dec 31 '18 at 17:29
  • Actualy i did (try), but since I don't understand his second step and did not get the answer, what could I do? @GitGud – nonuser Dec 31 '18 at 18:03