# Prime as sum of three numbers whose product is a cube

Good evening!
I am very new to this site. I would like to put the following material from Prof. Gandhi’s note book and my observations. Of course it is little long with more questions. But, with good belief on this site, I am sending for good solutions/answers.

If we take other than primes $2$, $5$ and $11$, every prime can be written as $x + y + z$, where $x$, $y$ and $z$ are some positive numbers. Interestingly, $x \times y \times z = c^3$, where $c$ is again some positive number.
Let us see the magic for primes $3,7,13,31,43,73$

Can you justify the above pattern? How to generalize the above statement either mathematically or by computer?

But, I observed that it is true for primes less than $9500$. Can your provide a computational algorithm to describe this?

Also, prove that, we conjecture that except $1, 2, 3, 5, 6, 7, 11, 13, 14, 15, 17, 22, 23$, every positive number can be written as a sum of four positive numbers and the product is again can be expressible in 4th power. Now, can we generalize this? Also, I want to know that, is there any such numbers can be expressible as some of $n$-integers with their product is again in $n$-th power?

Thank you so much.

edit

Concerning this cubic property :

Notice that this can be extended to hold for almost all squarefree positive integers $> 2$, not just the primes.

for instance :
we know for the prime $7$ : $7=1+2+4$ so we also get $7A = 1A + 2A + 4A$ and $1A * 2A * 4A$ is simply equal to $8A^3$.

In fact this can be extended to all odd positive integers $>11$ if $25,121$ have a solution.

Hence I am intrested in this and I placed a bounty.

I edited the question because its to much for a comment and certainly not an answer.

I would like to remind David Speyer’s comment : Every prime that is $1 mod 3$ is of the form $a^ 2 +ab+b^ 2$ , so that covers half the primes immediately.

So that might be a line of attack.

To add pessimism in finding such prime $p$ that cannot be written as sum of $3$ co-divisors of cube, I’ve drawn images:

Number of ways to write an integer number $p$ as the sum $x+y+z$, where $x\cdot y\cdot z=c^3$.

($\color{red}{\bf{red}}$ dots $-$ composite numbers, $\bf{black}$ dots $-$ prime numbers)

Control points:

$p=486$ (composite): $1$ way: $486 = 162+162+162$;

$p=2048$ (composite): $2$ ways: $2048 = 128+720+1200 = 224+256+1568$;

$p=6656$ (composite): $3$ ways: $\small {6656 = 416+2340+3900 = 512+1536+4608 = 728+832+5096}$;

$p=7559$ (prime): $4$ ways: $\scriptsize{7559 = 9+50+7500 = 114+225+7220 = 135+1024+6400 = 722+2809+4028}$;

$p=26624$ (composite): $5$ ways;

$p=58757$ (prime): $10$ ways;

$p=80429$ (prime): $13$ ways;

$p=111611$ (prime): $15$ ways;

$...$

These images were made in style like here.
(Since I’ve seen images for Goldbach’s conjecture, I lost any hope to find contradiction for it).

Method of search:

to build array $a[3N]$ for storing number of ways;

$a[p]$ is the number of ways to write $p$ as such sum; (initially, each $a[j]=0$);

for $c = 1 ... N$
$\quad$ create list of prime divisors of $c^3$;
$\quad$ (there is enough to know prime decomposition of $c$)
$\quad$ for each pair $x,y$ of divisors of $c^3$ to find $z=\dfrac{c^3}{xy}$;
$\quad$ if $z\in\mathbb{Z}$ and if $x\le y\le z$, then increase $a[x+y+z]$.

(if $c>N$, then sum of co-divisors of $c^3$ is greater than $3N$).