3

I suppose this property has it's name , so i apologize in advance for the ambiguous title. Suppose we are given a integer polynomial $P$ and three different arbitrary integers $a,b,c$ prove that the following is never true $$P(a)=b, P(b)=c, P(c)=a$$

My attempt: Well i have a feeling this is extremely simple but i just couldn't put my finger on it so i decided to try a not so strict method, which may not be correct. I mad an equation system as follows:

$$\alpha a^2+\beta a+\gamma=b$$ $$\alpha b^2+\beta b+\gamma=c$$ $$\alpha c^2+\beta c+\gamma=a$$

From here, I wanted to prove that this equation system has no solution or is impossible but even solving it is very complicated. I would like to know ( possibly just a hint) how to do this.

EDIT: We're dealing here with integer polynomial, not real polynomials. I apologize for such a stupid mistake,

Transcendental
  • 2,204
  • 1
    If the degree of the polynomial can be arbitrary, any finite collection of prescribed values $p(x_i)=y_i$ can be realized (see http://en.m.wikipedia.org/wiki/Lagrange_polynomial)! I think you are missing an assumption. – PhoemueX Feb 08 '15 at 10:06

2 Answers2

4

For an integer polynomial we have $(a-b) | (P(a)-P(b)) = (b-c)$. So we get (ideal notation)

$$(a-b) \supset (b-c) \supset (c-a) \supset (a-b)$$.

This shows $|a-b|=|b-c|=|c-a|$, which is obviously not possible for $3$ distinct integers.

MooS
  • 31,390
  • And in case OP is not familiar with ideals, what you get is $a-b$ divides $b-c$, which divides $c-a$, which divides $a-b$, from which it is not hard to derive a contradiction to the hypotheses. – Gerry Myerson Feb 08 '15 at 11:10
  • What I had in mind when choosing the ideal notation: $a-b|b-c|c-a|a-b$ looks pretty ugly. – MooS Feb 08 '15 at 11:20
3

Something is missing take $P(X)=-\frac{3}{2}X^2-\frac{1}{2}X+1$. We then have $P(-1)=0$, $P(0)=1$ and $P(1)=-1$

marwalix
  • 16,773