(In reply to
How about ... by Larry)
Assume a is even, say a=2k then N=3a^2-ab^2-2b-4 becomes 2*(6k^2-kb^2-b-2). This is clearly even so N!=1 in this case.
Assume b is even, say b=2m then N=3a^2-ab^2-2b-4 becomes 3a^2 - 4*(am^2-m-1). Take this mod 4, then we need 3a^2=1 mod 4 in order for N to equal 1. But 3a^2 can only be 0 or 3 mod 4. So N!=1 in this case.
So we are down to needing both a and b to be odd. Then N=3a^2-ab^2-2b-4 will have two odd terms (3a^2 and ab^2) and two even terms (2b and 4) so by parity N must again be even. Therefore N can never equal 1.
Actually I believe this is strong enough to say N cannot be congruent to 1 mod 4, so things like N=5 or 9 are also impossible.
re: How about ... a proof
