Conjecture:
Let $p$ be an even perfect number, and $a, b, c$ be positive natural numbers.
There exists values for $a, b, c$ to satisfy the following equation $$a^3 + b^3 + c^3 = p^3$$ for which $\bigg(\dfrac{a + b + c}{2}\bigg)^3 = x^3 + y^3 + z^3$ for natural numbers $x, y, z$.
Tested for all perfect numbers up to $2^{30}(2^{31} - 1) = 2305843008139952128$. I posted something similar relating to this a while ago on the Math Overflow and now I need better computer-power.
Ideally, I want a better way to approach this problem in order to prove/disprove this, but I do not know how or where to begin. Could somebody please help me?
Also, this is just a speculation, but every prime number can be written of the form $a^3 - b^3 - c^3$ for some natural $a, b, c$. This includes $2$, but there is already a formula for $2$: $$2 = (6t^3 + 1)^3 - (6t^3 - 1)^3 - (6t^2)^3$$
Also, if you have the following equation: $$\qquad \ \ \ a^3 + b^3 + c^3 = \big\{d^3 : \min\{a, b, c\} = 6\big\}$$ $$\implies 6 \mid d + (-1)^{n + 1}$$ for which $a^3 + b^3 + c^3 = d^3$ is the $n^{th}$ equation such that $\min\{a, b, c\} = 6$.
Thank you in advance.
Edit:
Here is the link to my post on the $MO \longrightarrow$ About prime numbers and the sum of three cubes
And here are a few examples of $a, b, c, x, y, z$ as part of the conjecture in the yellow box:
Let $a = 3, b = 4, c = 5$ then $x = 3, y = 4, z = 5$.
Let $a = 18, b = 19, c = 21$ then $x = 11, y = 15, z = 27$.
Let $a = 57, b = 82, c = 495$ then $x = 15, y = 213, z = 281$ or $x = 162, y = 173, z = 282$.
Let $a = 2979, b = 4005, c = 7642$, this was too high for me to calculate a while ago so I went here to find the value of $\big((a + b + c)/2\big)^3$ which I did, being equal to $7313$, meaning that $7313^3$ can be written as the sum of three positive cubes.
