# A polynomial determined by two values

Prove or disprove: a non constant polynomial $P$ with non-negative integer coefficients is uniquely determined by its values $P(2)$ and $P(P(2))$.

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True. If $P(x)=a_nx^n+\cdots+a_1x+a_0$ then each of the coefficients are less then $b\equiv P(2)$. Each of these coefficients can then be read off from the base-b expansion of $P(b)=P(P(2))$.

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Several people seem to be answering this one at the same time (how to select a single accepted answer?). Actually, it's a common question that keeps popping up on various maths forums. –  George Lowther Aug 26 '10 at 22:40
For what it's worth, Steve got there a few seconds before me. –  George Lowther Aug 26 '10 at 22:43
Perhaps you can wait for the 3rd upvote and then delete. You will get a badge :-) –  Aryabhata Aug 26 '10 at 22:48
Yeah, I saw that you deleted your answer, but I don't see any issue with having two the same. I just voted Steve up and pointed out that he did get there before before me in the comment, to be fair to him. –  George Lowther Aug 26 '10 at 22:50
Oh, I like badges :) –  George Lowther Aug 26 '10 at 22:51

Look at $P(P(2))$ in base $P(2)$. The nth place is the coefficient of $x^n$.

Steve

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You don't have to sign your answers, BTW. –  ShreevatsaR Aug 27 '10 at 5:57
I don't "have" to, I want to :). I do it on all the other forums, so force of habit. –  user641 Aug 27 '10 at 19:08