So I was given a challenge by my maths teacher and I do not know how to solve it:
Challenge:
Find three prime numbers $p, q, r$ such that $pqr(p + q + r) = P = \text{Perfect Number}$.
Ultimate:
Do not use a calculator or have any access to the internet.
My Attempt for the Ultimate Part of the Challenge:
Now so far, nobody has discovered an odd perfect number, thus $P$ is even, meaning that one of the prime numbers has to be $2$. I will denote $2$ as $p$: $$\therefore 2qr(2 + q + r) = P$$ Now I also know that primes are always odd (otherwise they will have a factor of $2$ so then it is not prime, excluding that $2$ is prime), so two primes when added together will equal an even number, and adding $2$ to that will also give an even number: $$\therefore 2 + q + r = \{2x : q \lor r \neq 2 \ \land \ x\in \mathbb{N} \because (2, q, r)\in \mathbb{N}\}$$ $$\therefore 4qrx = P$$ Since $4\mid P$ then this implies that $P\neq 6$ because $4\nmid 6$. For $28$, $28/4 = 7$ which is a prime number itself and has no (prime) divisors, which implies that $P\neq 28$. In fact, any perfect number is of the form $2^{n - 1}(2^n - 1)$, so this means that one of the numbers $2$, $q$, $r$ must equal each other since $qr \neq 2^n - 1$ since $2^n - 1$ is prime (a prime of this form is known as a Mersenne Prime and it's symbol is denoted as $M_p$). We know that: $$\begin{align} 2qr(2 + q + r) &= 4qr + 2q^2r + 2qr^2 \\ &= 4qr\bigg(1 + \frac{q}{2} + \frac{r}{2}\bigg) \\ &= 4qr\bigg(1 + \frac{q + r}{2}\bigg) \\ \therefore x &= 1 + \frac{q + r}{2} \end{align}$$ $$\because q \lor r \neq 2 \implies q + r = 2m + 1 : m\in \mathbb{N} \implies x\in \mathbb{N}$$ Here we can actually get $q$ and $r$ on both sides of the equation: $$\frac{P}{4qr} = x = 1 + \frac{q + r}{2}$$ Now since $P$ is of the form $2^{n - 1}(2^n - 1)$, we know that $2^n - 1$ is prime, so this means that: $$2 + q + r = 2^n - 1 \ \land \ 2qr = 2^{n - 1}$$ Now the fault in this is that $2^n - 1$ is odd, which means that $q \lor r = 2$, and thus means that $x\in \mathbb{Q}\notin \mathbb{N}$. Let $q = 2$: $$\therefore 4 + r = 2^n - 1 \ \land \ 4r = 2^{n - 1} \implies n > 3$$ This seems true because if $n = 3$ then substituting for $n$ in the expression $2^{n - 1}(2^n - 1)$, you get $28$ (and this was when I slapped myself in the head).
Now when I tested for $r = 3$, I got that $12 = 2^6$ although $12$ and $2^6$ have a drastic gap between each other since $2^6 = 4^3 = 64$. Also, $r \neq 2$ (and obviously $1$ since $1$ is not prime) because the next mersenne prime above $4$ is $7$, and $7 - 4 = 3$.
The next mersenne prime above $7$ is $31$, and $31 - 4 = 27$, but $r \neq 27 \because 27 = 3^3$ which is not prime.
The next mersenne prime above $31$ is $127$, and $127 - 4 = 123$, but $r \neq 123 \because 3\mid 123$ (I know from using the "divisibility by $3$ rule").
There seems to be a pattern here, where $r$ seems to always be divisible by $3$. I used the "divisibility by $3$ rule" on $8191 - 4$ and $131071 - 4$ and they seem to all be divisible by $3$. (I do not remember any other perfect numbers after that from the top of my head).
So it seems like there is no value for $r$ except $3$ itself, which is a contradiction to the fact that $pqr(p + q + r) = P$. So now I have two questions (and I am not taking the Ultimate challenge anymore since I am seeking help from the internet):
- Does $ \ \exists(p, q, r) : pqr(p + q + r) = P$?
- Does $ \ 3\mid (M_p - 4)$?
I would mostly appreciate if you answered the two questions, but I would also appreciate it if you just answered one of the questions. Although he first question is the challenge question however, I really want to know the answer to the second question. Also, did I do anything wrong in my attempt of trying to solving the challenge?
Thank you in advance.