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**PaulRS** For all parts, note that if $\displaystyle d|n$ then $\displaystyle \left(\tfrac{n}{d}\right)|n$, the idea would be the following:

For each divisor $\displaystyle d<\sqrt{n}$ there's a corresponding one $\displaystyle \tfrac{n}{d}>\sqrt{n}$, the only divisor that is not counted here is $\displaystyle d = \tfrac{n}{d}=\sqrt{n}$ in case of this being possible (that is, n is a square).

Thus if $\displaystyle n$ is not a square the divisors come in pairs, hence the number of divisors in that case is even, and in the case of n being a square it is odd (we add one).

For the second part use the above idea and note that $\displaystyle d\cdot \left(\tfrac{n}{d}\right)=n$ -and now try to use it to calculate the product of the divisors-