Show that the quotient $\frac{\sigma(p^3)}{\tau(p^3)}$ is an integer for $p$ prime

Question

Let p $\geq$ 3 be a prime. Show that the quotient $\frac{\sigma(p^3)}{\tau(p^3)}$ is an integer.

I know that I have to use the product formulas but not exactly sure how to go from there.

Answer 1

$$\tau(p^n)=\#\{\text{Divisors of $p^n$}\}=\#\{1,p,p^2,\ldots,p^n\}=n+1.$$ $$ \sigma(p^n)=\ \text{Sum of he divisors of $p^n$=}\sum_{i=0}^np^i=\frac{p^{n+1}-1}{p-1} $$ So, $$ \frac{\sigma(p^3)}{\tau(p^3)}=\frac{p^4-1}{4(p-1)}=\frac{(p^2+1)(p+1)}{4}, $$as $p\ge 3$ so it is odd hence $p^2+1$ and $p+1$ both are even and therefore their product is divisible by 4.