Understand the problem
Source of the problem
Start with hints
Let us denote the residue class of by . Show that, there exist non-zero in such that and .
From hint 2, show that . This means that (you need to convince yourself that the inverses exist). Now it becomes a matter of simply choosing .
Note that, cannot be zero. Given any , there exists exactly one non-zero such that is 0 modulo . Hence, in this case there are choices. For , this special is actually 0. Hence in this case there are choices. Thus, the total number of choices is . Adding to this the cases considered in hint 1, we get as the answer.
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