If $q^k n^2$ is an odd perfect number with special prime $q$, then $q \geq 17$.

(Note: The contents of this blog post are based from this MSE question.) The topic of odd perfect numbers likely needs no introduction. Let $N$ be an odd perfect number given in the so-called Eulerian form $N = q^k n^2$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

It is known that $n^2 - q^k$ is not a square.  Note that $n^2 - q^k \equiv 0 \pmod 4$ is true since $n$ odd implies that $n^2 \equiv 1 \pmod 4$ holds.  In particular, we may write $n^2 - q^k = 2^r t$, where $\gcd(2,t)=1$ if and only if $\nu_{2}(n^2 - q^k)=r$. (Note that $n^2 - q^k$ is not squarefree.) Now, assume that $2^r \parallel (n^2 - q^k)$. (That is, suppose that $n^2 - q^k$ is divisible by $2^r$ but not by $2^{r+1}$.) Then $\gcd(2,t)=1$, and we have $n^2 - q^k = 2^r t$, where $r=r(n,q,k)$ and $t=t(n,q,k)$ are unique. Now, suppose that $r \geq 3$ holds.   Therefore, since $n$ is odd implies that $n^2 \equiv 1 \pmod 8$ holds, and since $2^r t \equiv 0 \pmod 8$, it follows that $q^k \equiv 1 \pmod 8$. Hence, we infer that $q \equiv 1 \pmod 8$ holds, since $k \equiv 1 \pmod 4$ implies that $$q^k \equiv q \cdot (q^2)^{(k-1)/2} \equiv q\cdot{1} \pmod 8.$$ Furthermore, $q \equiv 1 \pmod 8$ rules out $q = 5$ and $q = 13$.  We conclude $q \equiv 1 \pmod {16}$, or in other words, $q^k \geq 17$. By the contrapositive, $q^k < 17$ implies $r = 2$. But we know that the biconditional $$(q^k < 17) \iff \Bigg((k = 1) \land \left(q \in \{5, 13\}\right)\Bigg).$$ The interested reader is then encouraged to read the following paper.