The Legend of AIME I #14

The timer was ticking: 25:01 ... 25:00 ... 24:59. I was about two and a half hours into the AIME test, having solved questions #1-8, 10, and 12. I had two options: either dive into a sequence of geometry bash problems or settle with the mythical #14 NT problem. I was driven to problem 14 as it offered the prospect of an easy point in the face of some intimidating geometry problems and a ticking timer. I took a deep breath and skimmed through the problem statement. It was something to do with the sigma function, $\sigma(a)$. Great, that was just what I wanted: a multiplicative function $\rightarrow$ primes. As Andrew Lin had said in my AIME class, the first instinct with this information should be to examine the prime powers by writing out the prime factorization: $$a=p_1^{e_1}\cdots p_d^{e_d}.$$ Okay, now what did I have to do - find the smallest $n$ so that $2021\;\vert\;\sigma(a^n)-1$. Immediately, I used my factorization to simplify this statement: $$\sigma(a^n)=\sigma(p_1^{ne_1}\cdots p_d^{ne_d})=\prod_{i=1}^{d}(1+p_i+\ldots+p_i^{ne_i})=\prod_{i=1}^{d}\Big(\frac{p_i^{ne_i+1}-1}{p_i-1}\Big),$$ $$\sigma(a^n)\equiv\prod_{i=1}^{d}\Big(\frac{p_i^{ne_i+1}-1}{p_i-1}\Big)\equiv 1 \pmod {2021}.$$ At first, that did not sound too bad, but then the implication struck me: this condition had to hold for all integers! Oh, so this had to hold for all powers of primes, implying that $$\frac{p^{na+1}-1}{p-1}\equiv 1 \pmod{2021}\Leftrightarrow \frac{p^{na+1}-1}{p-1}-1\equiv\frac{p(p^{na}-1)}{p-1}\equiv 0 \pmod{2021},$$ where $p$ could be any prime and $a\in\mathbb{Z_{>0}}$. Now, the path to the solution was crystal clear. I glanced at the timer to ensure that I was at pace with its sprinting digits: 14:17 ... 14:16. Wow! I was solving this problem as fast as the preliminary problems. Now, I set out to see when the divisbility constraint would hold.
Firstly, $p$ was never going to be divisible by $2021$ because primes did not have any factors besides $1$ and itself. So I rewrote the new condition in bold on the middle of my scratch paper: $$\mathbf{\frac{p^{na}-1}{p-1} \pmod{2021}}.$$ The biggest enemy in a modulo problem is the denominator. I had to find a way of getting rid of the denominator, so I gave into the dreaded idea of casework:
a. If $p-1\not\mid 43,47$, then I could ignore it completely (a mathematician's dream): $$\frac{p^{na}-1}{p-1} \equiv 0\pmod{2021}\Rightarrow p^{na}\equiv 1 \pmod{2021}.$$ A classic problem for Euler's theorem: $$p^{na}\equiv 1 \pmod{2021}\Leftrightarrow \varphi(2021)=42\cdot46\;\vert\;na.$$ Since $a$ could be any integer, the above condition eventaully reduces to $42\cdot46\;\vert\;n$.
b. If $p-1\;\vert\;43,47$, then I could use LTE. Sweet ;) $$\nu_{43}(p^{na}-1)=\nu_{43}(p-1)+\nu_{43}(na).$$ $$\nu_{47}(p^{na}-1)=\nu_{47}(p-1)+\nu_{47}(na).$$ For $p^{na}\equiv 1 \pmod{2021}$, we would need to have that $\nu_{43}(na)\geq 1$ and $\nu_{47}(na)\geq 1$. Again, $a$ could be any positive integer, so $43\cdot47\;\vert\;n$.
This was it: $42\cdot46\;\vert\;n$ and $43\cdot47\;\vert\;n$. The smallest $n$ that fit these conditions is $42\cdot43\cdot46\cdot47$. Hence, my final answer is $42+43+46+47=176$. I typed in my answer and checked back through my work. After skimming my scracth paper a few times, I made sure that I did not misbubble any of the previous answers.
I was now headed down the home stretch with one minute left on the exam time. It was only then that I realised that problem #14 asked me to find the sum of the distinct prime factors of $n$. $42$ and $46$ were not prime! I quickly factored $$n=2^2\cdot3\cdot7\cdot23\cdot43\cdot47 \text{ and got the right answer of } 2+3+7+23+43+47=\boxed{125}.$$ Phew! That was a narrow escape. 0:02 ... 0:01 ... 0:00. As the timer ran out, it not only marked the end of the AIME test but also the end of my suspicions about the myth of problem #14. Despite its easy outlook, the NT problem at that place in the AIME exam always has the knack of throwing off test-takers. I found it rather ironcial that I started the problem by instinctively thinking about primes and ended up nearly making a silly mistake by overlooking that very same idea of primes. Anyways, I scored 11/15 on the AIME test, and my experince with problem 14 makes this test all the more memorable.