b) In order a number $k$ to have 10 positive divisors, it must be either of the form $p^9$ or $pq^4$ where $p,q$ are primes. Let's assume there exist such $n$ having the property that $n^2-4$ has exactly 10 positive divisors. 1) $n^2-4=p^9$ Obviously, $n$ is odd because otherwise we have contradiction modulo 8. Then we have $(n+2)(n-2)=p^9$ and since $n$-odd we have $gcd(n+2,n-2)=1$. Therefore, we have $p^9=5$, which is absurd. 2) $n^2-4=pq^4$. Again $n$ must be odd. Then we have $(n+2)(n-2)=pq^4$ and $gcd(n+2,n-2)=1$ which implies one of the following cases: 2.1) $n-2=p, n+2=q^4\Rightarrow q^4-4=p\Rightarrow (q^2-2)(q^2+2)=p$, which is impossible since $q^2$ cannot be $3$. 2.2) $n-2=q^4, n+2=p\Rightarrow q^4+4=p$. Considering modulo 5 we see that if $q\not=5$ then $p=5$ - absurd. If $q=5$ we obtain $p=629=17.37$, which is not prime. 2.3) $n-2=1,n+2=pq^4\Rightarrow pq^4=3$ - contradiction.

broniran wrote: b) In order a number $k$ to have 10 positive divisors, it must be either of the form $p^9$ or $pq^4$ where $p,q$ are primes. Let's assume there exist such $n$ having the property that $n^2-4$ has exactly 10 positive divisors. 1) $n^2-4=p^9$ Obviously, $n$ is odd because otherwise we have contradiction modulo 8. Then we have $(n+2)(n-2)=p^9$ and since $n$-odd we have $gcd(n+2,n-2)=1$. Therefore, we have $p^9=5$, which is absurd. 2) $n^2-4=pq^4$. Again $n$ must be odd. Then we have $(n+2)(n-2)=pq^4$ and $gcd(n+2,n-2)=1$ which implies one of the following cases: 2.1) $n-2=p, n+2=q^4\Rightarrow q^4-4=p\Rightarrow (q^2-2)(q^2+2)=p$, which is impossible since $q^2$ cannot be $3$. 2.2) $n-2=q^4, n+2=p\Rightarrow q^4+4=p$. Considering modulo 5 we see that if $q\not=5$ then $p=5$ - absurd. If $q=5$ we obtain $p=629=17.37$, which is not prime. 2.3) $n-2=1,n+2=pq^4\Rightarrow pq^4=3$ - contradiction. Why would n can't have 1 divisor 2?

For part $a)$, the solution is missing (and, in all resources I've seen; the examples are stated, without any mention to motivation as to how to get those answers). So, I want to try to give a slightly motivated way of seeking an answer for the first part. Let us seek for, numbers of form, $pq^4$. For, $n^2-1$; this is really easy; and, let, $q=2;p=3$, and, $n=7$. It is little more delicate for the other two cases. Consider, first, $n^2-2=pq^4$; and suppose, $p,q>2$. Clearly, $n$ is odd; hence, $pq^4 \equiv -1 \pmod{4} \implies p\equiv -1 \pmod{4}$. Next, we also have, $$ n^2 \equiv 2 \pmod{p} \implies \left(\frac{2}{p}\right)=1. $$Hence, $p\equiv \pm 1 \pmod{8}$. Since, $p\equiv 3 \pmod{4}$; we get, $p\equiv 7 \pmod{8}$. Also, $(2/q)=1$; and this gives, $q\equiv \pm 1 \pmod{8}$. The smallest candidate, is, $q=7$; and, $p\equiv 7\pmod{8}$, $p>7$; thus, $p=23$. Trying this, we indeed get, $7^4 \cdot 23 = 235^2-2$, as desired. \noindent Now, let us study, $n^2-3=pq^4$. Suppose that, $p,q>2$; and, therefore, $n^2 \equiv 0 \pmod{4}\implies pq^4\equiv - 3\pmod{4}$. Hence, $p\equiv 1 \pmod{4}$. Also, $3$ is a quadratic residue, modulo $p$; and, modulo $q$. $$ \left(\frac{-3}{p}\right)=\left(\frac{3}{p}\right)\left(\frac{-1}{p}\right)=1; $$hence, $p\equiv 1\pmod{6}$. What this means, is, $p\equiv 1\pmod{12}$. Also, $(3/q)=1$; hence, depending on the value of $q$ in modulo $4$; we have, either, $$ q\equiv 1\pmod{12}\quad \text{or}\quad q\equiv 11\pmod{12}. $$Smallest candidates, therefore, are, $q=13,37,11,23$, etc. Using, $p=13$; and, $q=37$, gives us an example.