f(x) = (x + b)2 + c, where b and c are integers. If the prime p divides c, but p2 does not divide c, show that f(n) is not divisible by p2 for any integer n. If an odd prime q does not divide c, but divides f(n) for some n, show that for any r, we can find N such that qr divides f(N).
Solution
The first part is trivial. If p does not divide (x+b), then it does not divide (x+b)2, so it does not divide (x+b)2+c. On the other hand, if p does divide x+b, then p2 divides (x+b)2, so p2 does not divide (x+b)2+c.
For the second part, we use induction on r. For r = 1, we are given that q divides f(n). Now suppose that qr divides f(N) for some N. If qr+1 divides f(N), then we are done. So suppose qr+1 does not divide f(N), so f(N) = qrh where q does not divide h. We have f(N+kqr) = f(N) + qr(2N+2b)k = qrh + qr(2N+2b)k. Now q divides (N+b)2+c, and does not divide c, so it does not divide (N+b)2 and hence does not divide N+b. It is odd, so it does not divide 2N+2b. Hence we can find k such that k(2N+2b) = -h mod q. Then we have qr+1 divides f(N+kqr), which completes the induction.
© John Scholes
jscholes@kalva.demon.co.uk
26 January 2004
Last corrected/updated 26 Jan 04