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Copyright © 1996-2018 Alexander BogomolnyIn the following we'll assume f(x) is a polynomial with integral coefficients:

_{n}x

^{n}+ c

_{n-1}x

^{n-1}+ ... + c

_{1}x + c

_{0}

Also, the vertical bar (the pipe symbol) is used to indicate divisibility: a|b is a shorthand for "a divides b."

## Lemma

For any two different integers p and q,

## Proof

Indeed,

_{n}(p

^{n}- q

^{n}) + c

_{n-1}(p

^{n-1}- q

^{n-1}) + ... + c

_{1}(p - q)

and, since (p - q) | (p^{k} - q^{k}) for every integer k>0, Lemma follows.

Assume then that

with all a,b,c, and d different. From Lemma we immediately obtain that

(d - a) | (f(d) - f(a)) = 3 - 2 = 1 | , and |

(d - b) | (f(d) - f(b)) = 3 - 2 = 1 | , and |

(d - c) | (f(d) - f(c)) = 3 - 2 = 1 |

Thus differences d - a, d - b, d - c all divide 1. But 1 has only
two divisors: 1 and -1. Therefore, by the Pigeonhole Principle, two of the differences coincide. Which
contradicts our assumption that the numbers

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Copyright © 1996-2018 Alexander Bogomolny64640376 |