6th Mexico 1992 Problem A2 Given a prime number p, how many 4-tuples (a, b, c, d) of positive integers with 0 < a, b, c, d < p-1 satisfy ad = bc mod p?
Given a, b, c ≠ 0 mod p, we can always find d ≠ 0 mod p such that ad = bc mod p. However, it is possible that d = p-1 mod p. If bc = 1 mod p, then a(p-1) ≠ 1 mod p, so d is not p-1 for any choice of a. On the other hand if bc ≠ 1 mod p, then if a = -bc mod p we will get d = p-1 mod p. Thus we may choose b arbitrarily (p-2 choices). Then for c = 1/b (one choice), we have p-2 choices for a, and d is then determined. For c ≠ 1/b (p-3 choices), we have p-3 choices for a, and d is then determined. Thus in total there are (p-2)(p-2 + (p-3) 2 ) = (p-2)(p 2 -5p+7) possibilities. Thanks to Suat Namli 6th Mexican 1992 © John Scholes jscholes@kalva.demon.co.uk 28 February 2004 Last corrected/updated 28 Feb 04