2

Statement of the problem:

Consider polynomials $P, Q$ with natural coefficients (including zero) such that no coefficient of the polynomial $P$ is greater than $b-1 \in \mathbb{N}$. Given points $a, b\in \mathbb{Z}^{+}$ such that $a<b$ and equivalences $P(a)=Q(a)$, $P(b)=Q(b)$, prove that $P$ is the same polynomial as $Q$.

Observations:

Let the coefficients of $P$ be $p_0,p_1, ...$ and $Q$ be $q_0,q_1,\dots$. Under these circumstances, the polynomial $P-Q$ must contain the factor $(x-a)(x-b)$, so the zeroth order term $p_0-q_0$ must be a multiple of $ab$.

Also, if $\lambda(x) = \frac{P(x)-Q(x)}{(x-a)(x-b)}$, the highest degree term has the coefficient $p_n-q_n$, where $n = \max(\deg(Q), \deg(P))$. This becomes evident with long division.

pjq42
  • 707
  • What about $P = x^3 + 2$, $Q = 2 x^2 + 1$, $a = 1$, $b = 2$? – Robert Israel Apr 23 '23 at 16:05
  • Note that if the coefficients could only go up to $b-1$, $P(b)$ and $Q(b)$ would then be writing a number in base $b$. Since there is only one unique to write a natural number in a given base, we could conclude that the coefficients of $P$ and $Q$ would be the same and therefore $P=Q$. Perhaps you can modify this argument, where knowing that $P(a)=Q(a)$ compensates for the fact that the coefficients range up to $b$ instead of $b-1$ . – Anton V. Apr 23 '23 at 16:10
  • @RobertIsrael Correct! I read the problem wrong. It actually requires that the coefficients of $P$ be explicitly less than $b$. I have edited it now. – pjq42 Apr 23 '23 at 16:47
  • @AntonV. Quite a nice idea, but since $Q$'s coefficients don't have any restriction, I still don't see a way. – pjq42 Apr 23 '23 at 16:49
  • @RobertIsrael $P(b)$ is not equal to $Q(b)$ in your example? – Anton V. Apr 23 '23 at 17:00
  • @pjq42 Oh yeah sorry, I misread the original problem... – Anton V. Apr 23 '23 at 17:46
  • @AntonV. Sorry, that should have been $P = x^3+2$, $Q = 2 x^2 + x$. – Robert Israel Apr 24 '23 at 05:13

0 Answers0