Solution by Doyle Henderson, Omaha, NE. The only possibilities for m are 0, 1, 2, 3, and 5. Simple calculations then show the solution set for (m, n) to be {(0, 0), (1, 1), (2, 2), (5, 11)1}. Suppose there is a solution with m even and at least 4. Let m = 2a and x = 3a We have x2 2n2 = 1. Since this is a Pell equation, it follows that x = Pj for some nonnegative integer j, where Po = 1, P1 = 3, and Pk = 2Pk-1 + Pk-2 for k > 2. Since a > 2, it follows that 9 divides Pp. Since Pk+12 = 9 11 140Pk+1 + 5741Pk, and P5 is the only element of {Po, ... , P11} that is divisible by 9, it follows inductively that j = 5 mod 12. Induction then shows that 11 divides Pj, which is a contradiction. Now suppose there is a solution with m odd and at least 7. Let m = 2b + 1 and y = 3b and z = 2n. We have Z2 6y2 = -2. This is another Pell equation, so y = Qj for some nonnegative integer j, where Qo = 1, Q1 = 2, Q2k = 4Q2k-1 ? Q2k-2, and Q2k+1 = 2Q2k + Q2k-1 for k > 1. Since b > 3, we find that 27 divides Qj. Since Q2k+9 = 27 .17 19Q2k+1 + 1960Q2k, and Q2k+10 = 27 .17 19Q2k+2 + 3920Q2k+1, and Q8 is the only element of { Qo, Q1 ... Q8} divisible by 27, induction shows that j 8 mod 9. It then follows inductively that 17 divides Q j, which is a contradiction.