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.

Solution 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, f(p) - f(q) is divisible by p - 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 a, b, c are all different. • 17 rooks on a Chessboard
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