Suppose f(x) is a polynomial with integral coefficients. If f(x) = 2 for three different integers a,b, and c, prove that, for no integer, f(x) can be equal to 3.
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Copyright © 1996-2018 Alexander Bogomolny
Suppose f(x) is a polynomial with integral coefficients. If f(x) = 2 for three different integers a,b, and c, prove that, for no integer, f(x) can be equal to 3.
In the following we'll assume f(x) is a polynomial with integral coefficients:
f(x) = cnxn + cn-1xn-1 + ... + c1x + c0
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,
(p - q) | (f(p) - f(q))
Proof
Indeed,
f(p) - f(q) = cn(pn - qn) + cn-1(pn-1 - qn-1) + ... + c1(p - q)
and, since (p - q) | (pk - qk) for every integer k>0, Lemma follows.
Assume then that
f(a) = f(b) = f(c) = 2 and f(d) = 3
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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