Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

From a St. Petersburg school olympiad, 11th grade.

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

share|cite|improve this question
up vote 13 down vote accepted

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 $P(b)=P(P(2))$.

share|cite|improve this answer
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 $x^n$.


share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.