p is an odd prime congruent to 2 mod 3. Prove that at most p-1 members of the set {m2 - n3 - 1: 0 < m, n < p} are divisible by p.
Solution
We show that p has p distinct cubic residues if it is an odd prime congruent to 2 mod 3. Let k be a primitive root of p. So any non-zero residue of p can be written as kn for some n in the range 1, 2, ... , p-1, and kn = km mod p iff p-1 divides m-n. Now we claim that, for distinct m, n in the range 1, 2, ... , p-1, we have k3n and k3m distinct mod p. For if they are equal then p-1 divides 3(m-n). But p-1 is not a multiple of 3, so p-1 must divide m-n. But m-n is in the range 1, ... , p-2 so it cannot be divisible by p-1.
So the values -13, -23, ... , -(p-1)3 are all incongruent mod p. Thus for each m, the p-1 values m2 - 13 - 1, m2 - 23 - 1, ... , m2 - (p-1)3 - 1 are all incongruent mod p. Hence at most one of them is divisible by p. So the set {m2 - n3 - 1: 0 < m, n < p} has at most one member divisible by p for each m and hence at most p-1 in all.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002