A prime p has decimal digits pnpn-1...p0 with pn > 1. Show that the polynomial pnxn + pn-1xn-1 + ... + p1x + p0 has no factors which are polynomials with integer coefficients and degree strictly between 0 and n.
Solution
Solution by Demetres Christofides, who also corrected the question
If w is a (complex) root of the polynomial, then since each coefficient is at most 9, we have |pn| ≤ 9(1/|w| + 1/|w|2 + ... + 1/|w|n-1). So if |w| ≥ 9, then |pn| ≤ 9(1/9 + 1/81 + ... ) < 9/8. But pn ≥ 2, so we must have |w| < 9.
Suppose the polynomial has a factor with integer coefficients (and positive degree less than n). Then Gauss' lemma tells us that it must be a product of two polynomials with integer coefficients (and each with positive degree less than n). Suppose they are f(x) and g(x). Suppose the (complex) roots of f(x) are z1, ... , zm. Then f(x) = A(x - z1) ... (x - zm) with A and integer. Hence |f(10)| = |A| |10 - z1| ... |10 - zm|. But |A| ≥ 1 and each factor |10 - zi| > 10 - 9 = 1. So |f(10)| > 1. Similarly, |g(10)| > 1. But f(10) and g(10) are integers and their product is the prime p. So we have a contradiction. So the polynomial cannot have any such factor.
© John Scholes
jscholes@kalva.demon.co.uk
7 Sep 2002