some things to help you narrow down the search:

consider first the group of units of Z_{30}= {1,7,11,13,17,19,23,29}.

a group can only have ONE idempotent, its identity. for if a*a = a, then:

a^{-1}*(a*a) = a^{-1}*a

(a^{-1}*a)*a = e

e*a = e

a = e.

so 1 is the only idempotent that is also a unit.

this means any other idempotents must lie in the set {0,2,3,4,5,6,8,9,10,12,14,15,16,18,20,21,22,24,25, 26,27,28}.

clearly 0 is an idempotent, and 2,3,4, and 5 are not. we could try every one of these, manually checking to see if they are idempotent. is there a better way?

if a is idempotent, a^{2}= a.

that is: a(a - 1) = 0 (mod 30). let's think about this. 30 = 2*3*5.

so if a is a zero-divisor, it contains 2,3 or 5 in its factorization. take 8, 8 has the factor 2.

for 8*7 = 0 (mod 30) to be true, 7 would have to have a factor of 3 and 5 (so that the product is a multiple of 2,3 and 5).

but 7 does not have a factor of 3 OR 5. 8 is not idempotent. (and 8*8 = 4).

9 has 3 as a factor, but not 2 or 5. 9-1 = 8, and 8 does not have a factor of 5 (it does have 2), so 9*8 cannot possibly be a multiple of 30. 9 is not idempotent.

10 has 2 and 5 as a factor, and 9 (10 - 1) has 3 as a factor. so 10 is idempotent. see how this works?

use this reasoning to show 1,6,10,15 and 16 are all idempotent (and there are no other idempotents < 17). what is the next one?

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can any unit be nilpotent? why, or why not?

can any idempotent be nilpotent? why or why not? (careful, this is a trick question).

are the remaining elements necessarily nilpotent? (can 30 divide a power of a if a does not have 2,3 and 5 as a factor?).