Let f(x) = xn + 5xn-1 + 3, where n > 1 is an integer. Prove that f(x) cannot be expressed as the product of two non-constant polynomials with integer coefficients.
Solution
Suppose f(x) = (xr + ar-1xr-1 + ... + a1x ± 3)(xs + bs-1xs-1 + ... + b1x ± 1). We show that all the a's are divisible by 3 and use that to establish a contradiction.
First, r and s must be greater than 1. For if r = 1, then ± 3 is a root, so if n is even, we would have 0 = 3n ± 5·3n-1 + 3 = 3n-1( 3 ± 5) + 3, which is false since 3 ± 5 = 8 or -2. Similarly if n is odd we would have 0 = 3n-1(±3 + 5) + 3, which is false since ±3 + 5 = 8 or 2. If s = 1, then ±1 is a root and we obtain a contradiction in the same way.
So r ≤ n - 2, and hence the coefficients of x, x2, ... , xr are all zero. Since the coefficient of x is zero, we have: a1 ± 3b1 = 0, so a1 is divisible by 3. We can now proceed by induction. Assume a1, ... , at are all divisible by 3. Then consider the coefficient of xt+1. If s-1 ≥ t+1, then at+1 = linear combination of a1, ... , at ± 3bt+1. If s-1 < t+1, then at+1 = linear combination of some or all of a1, ... , at. Either way, at+1 is divisible by 3. So considering the coefficients of x, x2, ... , xr-1 gives us that all the a's are multiples of 3. Now consider the coefficient of xr, which is also zero. It is a sum of terms which are multiples of 3 plus ±1, so it is not zero. Contradiction. Hence the factorization is not possible.
Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
12 Nov 1998
Last corrected/updated 25 Aug 03