(mn-1)|(n³-1) Of course we must have n≧2 in all cases If m=1, we have (n-1)|(n³-1) which of course is true, since n³-1 = (n-1)(n²+n+1) If m=n², we have (n³-1)|(n³-1) which of course is true, since any natural number divides itself. Now if n is a perfect square, say k², then we have (mk²-1)|[(k²)³-1] (mk²-1)|(k6-1) Then we can have m=k, for then we'd have (k³-1)|(k3-1)(k3+1) So apparently there are 3 cases of solutions for n≧2 m=1, m=n², and the third case is when n is a perfect square, k², and m is its square root, k. Edwin