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Let $a$, $b$ and $c$ be three distinct integers and $P$ a polynomial with integer coefficients. Show that the conditions $P(a)=b, P(b)=c$ and $P(c)=a$ cannot be satisfied simultaneously.

Using polynomial remainder theorem:

  1. Remainder of the division of a polynomial $P(a)$ by a polynomial $a-b$ is equal to $P(b)$.

  2. Remainder of the division of a polynomial $P(b)$ by a polynomial $b-c$ is equal to $P(c)$.

  3. Remainder of the division of a polynomial $P(c)$ by a polynomial $c-a$ is equal to $P(a)$.

I can't find example where $P(a)=b,P(b)=c,P(c)=a$.

How to prove this contradiction?

user300045
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2 Answers2

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If $P(a)=b$ and $P(b)=c$ then we have $P(a)-P(b)=b-c$ which means that $a-b|b-c$. Using similar argument we can say that $a-c|b-a$ and $b-c|c-a$. But from $a-b|b-c$ we can say that $a-b|a-c$, that jointly with $a-c|b-a$ yields $|a-b|=|a-c|$. Similarly we have $|b-a|=|b-c|$ which means that: $$ |b-a|=|b-c|=|a-c|. $$ This means $a=b=c$ hence contradiction.

Arash
  • 11,307
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We suppose the conditions are satisfied. The polynomial $P - c$ has root $b$, so there exists a polynomial $Q_1$ such that $$ P(x) - c = (x - b) Q_1(x). \qquad (1) $$ Similarly, there exist polynomials $Q_1$ and $Q_2$ such that $$ \begin{align*} P(x) - b &= (x - a) Q_2(x) \\ P(x) - a &= (x - c) Q_3(x). \end{align*} $$ Take the largest number among $\lvert a-b \rvert$, $\lvert a - c\rvert$ and $\lvert b - c \rvert$. We suppose it is $\lvert a - b\rvert$. Therefore, by triangle inequality, we have $$ \lvert b - c\rvert \leqslant \lvert b -a \rvert + \lvert a - c \rvert < \lvert b - a \rvert \qquad (2) $$ because $a \neq c$. Substituting $x = a$ in $(1)$ gives $$ b - c = (a - b)Q_1(a). $$ Given that $Q_1$ is a polynomial with integer coefficients, $Q_1(a)$ is a non-zero integer (because $b\neq c$), so $$ \lvert b - c \rvert = \lvert a -b \rvert \times \lvert Q_1(a) \rvert \geqslant \lvert a - b \rvert. $$ This contradicts $(2)$.

Éric Guirbal
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