(not sure what the latex symbol is for equivalent)
so n=1 or -2 which is equivalent to 1 and 40 in mod 42.
Unfortunately, Deadstar and TheEmptySet miss two solutions:
If 2, 3, 7 all divide or , the solution set is
(i) If 2, 3 divide and 7 divides then The Chinese-remainder method (or otherwise trial and error) gives as one solution. Hence the solution set is
(ii) If 3 divides and 2, 7 divide then In this case, but we already have this solution above.
(iii) If 2 divides and 3, 7 divide then which gives the same solution set as in (i).
(iv) If 2, 7 divide and 3 divides then So which we already have above.
(v) If 3, 7 divide and 2 divides then So and we have found a new solution set.
(vi) Finally, if 7 divides and 2, 3 divide then and again we have
Hence the complete solution set is