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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.

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    $\begingroup$ Your conjecture would be wrong if $p$ were an odd perfect number, because $((a+b+c)/2)^3 would not be an integer. Do you want to restrict it to even perfect numbers? $\endgroup$ – Robert Israel Oct 27 '17 at 0:40
  • $\begingroup$ Oh yes thanks for the pick up. I will include that in my post $\endgroup$ – Mr Pie Oct 27 '17 at 1:05
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    $\begingroup$ A couple questions: 1. Can you please provide the link to your Math Overflow question if it is relavent? and 2. What are some of the values of $a,b,c,x,y,z$ that you have found to satisfy this? Since even perfect numbers satisfy a very nice form, I wouldn't be surprised if there exists a closed form for the choices of $a,b,c$. $\endgroup$ – Carl Schildkraut Oct 27 '17 at 1:26
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    $\begingroup$ A similar question has been discussed in this post. $\endgroup$ – Tito Piezas III Nov 10 '17 at 17:19
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    $\begingroup$ Your speculation about $a^3\pm b^3 \pm c^3 = N$ for prime $N$ is discussed for general $N$ here. $\endgroup$ – Tito Piezas III Nov 11 '17 at 3:59

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