hey, how's this - I think I have an answer..
does the prime factorization of n have to contain a repeating factor?
this makes sense to me because, for example if n = 18 then..
18 = (2)(3)(3) <---- there are two 3's here (3 is a repeating factor)
so if I divide by the repeating...
okay, so let's see..
you're asking me whether n itself is a nilpotent element of Z/nZ.
yes, I think n = (p_1)(p_2)...(p_r) is nilpotent by definition because:
n^k is congruent to 0 (mod n), for k > 0 ...right?
I'm not sure how your hint regarding the prime factorization can...
Could someone please tell me what condition on n is necessary and sufficient such that Z/nZ contains a nilpotent element?
I can't seem to find this in any of my texts.
By the way, I accidently posted this same question in the homework help section first. Sorry to anyone who's wondering...