Prove that if $n$ is a positive integer, there exists only one polynomial $\displaystyle P(x)=\sum_{i=0}^n a_ix^i$ degree $n$ that satisfies:



  • $\begingroup$ Homework? Really? For what class? $\endgroup$ Jan 25, 2014 at 14:40
  • 2
    $\begingroup$ Some ideas: (i) $a_k$ are the digits of $P(10)$ in base 10. (ii) $P(-2) \equiv a_0 \textrm{ (mod 2)}$ and $P(-5) \equiv a_0 \textrm{ (mod 5)}$, which imply $n \equiv a_0 \textrm{ (mod 10)}$. (iii) These seem to scream "Chinese remainder theorem" and "induction" but after a quick glance I didn't find anything straightforward. $\endgroup$
    – JiK
    Jan 25, 2014 at 15:05

1 Answer 1


Suppose there are two polynomials $P(x)$ and $Q(x)$ that satisfy the constraints. Then, we can take $R(x)=P(x)-Q(x)$. We know that $R(-2)=R(-5)=n-n=0$. Also, the coefficients of $R$ are in $\{-9,-8,\dots,8,9\}$. It follows that $$ R(x)=(x+2)(x+5)S(x)=(10+7x+x^2)S(x) $$ where $S(x)$ is the quotient of $R(x)$ when divided by $(x+2)(x+5)$. Now say that $$ S(x)=\sum_{i=0}^{n-2}b_ix^i $$ with $b_i$ integers, because $(x+2)(x+5)$ is monic. (We sum up to $n-2$, because we multiply with a polynomial of degree $2$ to get $R$.) Let $j$ be the smallest value of $i$ for which $b_i$ is not $0$, assuming such an $i$ exists. If it does not exist, we know that $S(x)=0$ for all $x$, so $R(x)$ is zero, which yields a contradiction, because $R(-2)=n\neq 0$. Now look at $a_j$. We know it has to be equal to $10 b_j$, because the only possible way to get a term with $x^j$ in $R$ is to take the $x^j$ from $S$ and the $10$ from $10+7x+x^2$ (because $x^{j-1}$ and $x^{j-2}$ don't occur in $S$). We find that $a_j=10 b_j$. Because $-9\leq a_j\leq 9$ and $b_j$ is an integer, $a_j=b_j=0$, but we assumed $b_j\neq 0$. Thus, we arrive at a contradiction, and thus, there can be only one polynomial with the required properties.

  • $\begingroup$ You don't need (and don't have) that $R$ is monic, it is enough that $(x+2)(x+5)$ is monic. So you don't need $P,Q$ monic either. Also, the fact that some $b_i\ne0$ is not from $R(-2)=n$ (after all $R(-2)=0$), but from $P\ne Q$. $\endgroup$ Jan 25, 2014 at 15:25
  • $\begingroup$ @HagenvonEitzen You're right, I'll edit it. $\endgroup$
    – Ragnar
    Jan 25, 2014 at 15:26

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