# 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 materiel 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 $3$, $7,13,17,\ldots$ \begin{align} 3 = 1 + 1 + 1 &\Longrightarrow 1 \times 1 \times 1 = 1^3\\ 7 = 1 + 2 + 4 &\Longrightarrow 1 \times 2 \times 4 = 8 = 2^3\\ 13 = 1 + 3 + 9 &\Longrightarrow 1 \times 3 \times 9 = 3^3\\ 17 = 1 + 8 + 8 &\Longrightarrow 1 \times 8 \times 8 = 4^3 \end{align} 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.

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I think you probably want a "such that" - something like "every prime other than $2$, $5$ or $11$ can be written as $x+y+z$ such that $xyz$ is a cube". Every positive integer greater than $2$ can be written as a sum of three positive integers, so you need some property of this triple to make the statement interesting. –  Matt Pressland Jun 15 '12 at 10:28
A further generalization would be that there is an integer $N_{k, m}$ such that for any prime $p\gt N_{k,m}$, there exist positive $x_1,\dots, x_k$ such that $x_1+\cdots+x_k=p$ and $x_1x_2\cdots x_k$ is a perfect $m$-th power. There is also no real reason to confine attention to primes. –  André Nicolas Jun 18 '12 at 15:39
There is no reason to focus attention on primes. Let's say that $N$ is clean if we can find positive integers $(x,y,z)$ with $x+y+z=N$ and $xyz$ a cube. If $N$ is clean, so is every multiple of $N$. Now, $2^6=7+8+49$, $5^2=3+4+18$ and $11^2=1+45+75$ are all clean. So the conjecture is equivalent to the much more natural conjecture every sufficiently large integer is clean. –  David Speyer Jun 18 '12 at 19:58
We can pull a little more information out. $\gcd(x,y,z) = 1$, otherwise $p$ wouldn't be prime. Let $\displaystyle x=\prod_{i=1}^{j}p_{i}^{a_{i}}$, $\displaystyle y=\prod_{i=1}^{k}p_{i}^{b_{i}}$, and $\displaystyle z=\prod_{i=1}^{l}p_{i}^{c_{i}}$. Let $m=\max(j,k,l)$. Then $\displaystyle xyz=\prod_{i=1}^{m}p_{i}^{a_{i}+b_{i}+c_{i}}$. If $xyz$ is a cube, then for all $1 \le i \le m$, we know that $a_{i}+b_{i}+c_{i} \in 3\mathbb{Z}$. But we also know that $gcd(x,y,z)=1$, so either $a_{i}=0$, $b_{i}=0$, or $c_{i}=0$. Suppose $c_{i}=0$. Then $a_{i}+b_{i} \in 3\mathbb{Z}$, so $(a_{i},b_{i})$ is in –  Jackson Walters Jun 18 '12 at 21:33
Every prime that is $1 \mod 3$ is of the form $a^2+ab+b^2$, so that covers half the primes immediately. –  David Speyer Jun 18 '12 at 22:15
I wrote a small script in MATLAB and verified your claim for all the primes less than $150,000$. Only $2,5,11$ do not satisfy your claim. For many of the primes, there are multiple ways to write it as a sum of three number such that its product is a cube. My script just looks for the first occurrence of such triplets for each prime.
Here is a .txt file with the "first" set of triplets for primes less than $150,000$.
I wrote some Mathematica code to compute the number of triples satisfying the property for a given $n$, with the following caveat. Suppose $(x,y,z)$ is a 3-tuple satisfying the property. Any permutation of $(x,y,z)$ also satisfies the property. Say $(x,y,z) \sim (x',y',z')$ if $(x',y',z')$ is a permutation of $(x,y,z)$. Then let $T_{n}$ be the number of such equivalence classes. Here's a plot of $T_{n}$ for $1 \le n \le 200$. Note: There are no solutions for $n= 44$ or $176$. picasaweb.google.com/lh/photo/… –  Jackson Walters Jun 19 '12 at 20:38