So I have made a theorem:
You are given two prime numbers $P_n$ and $P_{n + 1}$ such that $P_n$ denotes the $n^{th}$ prime number and $P_{n + 1}$ is the following prime number from $P_n$.
Take the product of $P_n$ and $P_{n + 1}$ and let that equal $d$. Let $σ(d)$ denote the sum of the digits of $d$ (even though generally $σ(d)$ denotes the sum of divisors of $d$ but we will ignore that for now).
$$σ(d) = 2r \iff \nexists m^2 \in (P_n, \ P_{n + 1})$$ This is read as
"The sum of the digits of $d = 2r$ for some $r \in \mathbb{N}$ if and only if (iff) there does not exist $m^2$ for some $m \in \mathbb{N}_{>2}$ between $P_n$ and $P_{n + 1}$ ($P_n$ and $P_{n + 1}$ not included)."
$$\text{If } \ \exists m^2 \in (P_n, \ P_{n + 1}) \implies σ(d) = 2r + 1$$. $$\text{but if } \ \frac{P_n + P_{n + 1}}{3} = \left\{\mathcal{P} : \mathcal{P} = \text{Perfect Number}\right\}$$ $$\implies \text{Theorem does not prove true for $d$ in this particular case.}$$
I do not know if I made that formula correctly. I could be completely off, but I got my information of how to write "in between" in mathematics from here.
Initially the bottom half of the theorem did not exist, but I did find counter-examples of the theorem such as $7\times 11$ and $41\times 43$.
Between $7$ and $11$, there does exist a squared number $m^2$ such that $m = 3$, but $σ(d) = 2r$ and not $2r + 1$ such that $d = 77$ and $r = 7$.
Between $41$ and $43$, there does not exist a squared number $m^2$, but $σ(d) = 2r + 1$ and not $2r$ such that $d = 1763$ and $r = 8$.
However with each of these counter-examples, I found one thing in common with them that makes them different from all the other examples (apart from their converse). $$\frac{7 + 11}{3} = \frac{18}{3} = 6$$ $$\frac{41 + 43}{3} = \frac{84}{3} = 28$$ $6$ and $28$ are both perfect numbers! (This discovery could potentially justify why perfect numbers end in either the digits $6$ or $8$, but that is for a seperate post.)
The reason $m \in \mathbb{N}_{>2}$ is because I only found one counter-example if $m \in \mathbb{N}$ instead $\longrightarrow 3\times 5$.
I guess my theorem is not really a theorem, but a mere speculation because I do not have a real proof or contradiction to this. How can I prove/disprove this; how must I approach this problem?
Thank you in advance.
Edit:
Doing some more testing, I believe there are infinitely many counter-examples, but they do not seem to be random.
It seems like if you have two prime numbers $P_n$ and $P_{n + 1}$ then this theorem does not prove true for $d$ in this other particular case if: $$\text{For some } k\in \mathbb{N}, \ \exists (3k)^6 \in (P_n, \ P_{n + 1})$$ Of course, $(3k)^6 = (3k)^{3\times 2} = \big((3k)^3\big)^2$ and so is a squared number.
A counter-example is if $P_n = 727$ and $P_{n + 1} = 733$. $729$ lies between $727$ and $733$ and is equal to $3^6$, and $σ(729\times 733) = 28$.
Another counter-example is if $P_n = 46649$ and $P_{n + 1} = 46663$. $46656$ lies between $46649$ and $46663$ and is equal to $6^6$, and $σ(46649\times 46663) = 50$.
However, if you have $9^6 = 531441$, a counter-example is not found, so $k\neq 3$. If $P_n = 531383$ and $P_{n + 1} = 531457$ where $531441$ lies between $531383$ and $531457$, $σ(531383\times 531457) = 35$ which is odd anyway $-$ not even!
This could be added to my speculation, since none of $P_n$ and $P_{n + 1}$ in these counter-examples give solutions to the equation $$\frac{P_n + P_{n + 1}}{3} = \mathcal{P}$$