clearly, we must have a subgroup isomorphic to (Z_{2},+), so n must be even. so if n = 2k, then this subgrioup is {[0],[k]}.

for this to be a subring of Z_{n}isomorphic to Z_{2}, we must have [k]*[k] = [k], that is:

k^{2}= k (mod n).

note that k^{2}= k + k +...+ k (k summands), if k is even (that is, n is divisible by 4) then this sum is 0 (mod n).

hence k must be odd. that is, n must be of the form n = 4t-2, for some positive integer t.

on the other hand, suppose n = 4t-2, and consider {[0],[2t-1]} in Z_{n}. we have:

[2t-1]*[2t-1] = [4t^{2}-4t+1] = [4t^{2}-2t] - [2t-1] = [4t-2]*[1] - [2t-1] = [0]*[1] - [2t-1] = -[2t-1] = -[2t-1] + [0] = -[2t-1] + [4t-2] = [2t-1].

note this shows that even in a ring with identity, a subring may have a different identity than the parent ring (something that is NOT true of groups).

some examples to consider: {[0],[3]} in Z_{6}, and {[0],[15]} in Z_{30}.