QuoteOriginally posted by: mattcushmanUse the fact that, mod 13, the integers are a field with the multiplicative group being cyclic of order 12. Thus, the cubes yield 4 distinct elements of Z/(13). Add in 0, and you get 5 distinct powers of 3 in Z/(13). So, defined x~y to mean X^3 = Y^3 (mod 13).Explicitly, these equivalence classes are given by 0, 1, 5, 8 and 12.Perfect! And even though the name of one of the equivalence classes is , I don't think that breaks the rule of no 5s since your relation itself doesn't contain 5. -t.