Polynomial challenge: Show that not all the coefficients of f(x) are integers.

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Discussion Overview

The discussion revolves around a degree 10 polynomial \( f(x) \) defined by the conditions \( f(p)=q \), \( f(q)=r \), and \( f(r)=p \), where \( p \), \( q \), and \( r \) are integers with \( p

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • Some participants propose that if all coefficients of \( f(x) \) are integers, then certain divisibility conditions must hold among \( p \), \( q \), and \( r \).
  • Others argue that the conditions imposed by the polynomial's behavior lead to contradictions if all coefficients are assumed to be integers.
  • A later reply corrects a minor typo in the divisibility condition mentioned, indicating a focus on the accuracy of mathematical statements.

Areas of Agreement / Disagreement

Participants appear to be engaged in a debate regarding the implications of integer coefficients, with no consensus reached on the validity of the initial assumptions or the conclusions drawn from them.

Contextual Notes

Limitations include potential missing assumptions about the relationships between \( p \), \( q \), and \( r \), as well as unresolved mathematical steps related to the implications of the polynomial's integer coefficients.

castor28
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$f(x)$ is a degree 10 polynomial such that $f(p)=q$, $f(q)=r$, $f(r)=p$, where $p$, $q$, $r$ are integers with $p<q<r$.

Show that not all the coefficients of $f(x)$ are integers.
 
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castor28 said:
$f(x)$ is a degree 10 polynomial such that $f(p)=q$, $f(q)=r$, $f(r)=p$, where $p$, $q$, $r$ are integers with $p<q<r$.

Show that not all the coefficients of $f(x)$ are integers.

Let all coefficients be integers
we have m-n divides $f(m)-f(n)$
so p- q | f(p) - f(q) | q-r

similarly
q -r | r-p
and r-p | p - q

from above as p - q | q-r | r- p | p- q (all nteger multiples) so all are same and hence a contradiction

so all coefficients cannot be integers
 
Last edited:
kaliprasad said:
Let all coefficients be integers
we have m-n divides $f(m)-f(n)$
so p- q | f(p) - f(q) | q-r

similarly
q -r | r-p
and r-q | p - q

from above as p - q | q-r | r- p | p- q (all nteger multiples) so all are same and hence a contradiction

so all coefficients cannot be integers
That is quite correct (except for a small typo: $r-q\mid p-q$ should be $r-p\mid p-q$).
Congratulations!:)
 
castor28 said:
That is quite correct (except for a small typo: $r-q\mid p-q$ should be $r-p\mid p-q$).
Congratulations!:)

Thanks castor. Corrected the same inline for the flow.
 

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