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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 $3,7,13,31,43,73$ $$ \begin{align} 3 = 1 + 1 + 1 &\Longrightarrow 1 \times 1 \times 1 = 1^3\\ 7 = 1 + 2 + 4 &\Longrightarrow 1 \times 2 \times 4 = 2^3\\ 13 = 1 + 3 + 9 &\Longrightarrow 1 \times 3 \times 9 = 3^3\\ 31 = 1 + 5 + 25 &\Longrightarrow 1 \times 5 \times 25 = 5^3\\ 43 = 1 + 6 + 36 &\Longrightarrow 1 \times 6 \times 36 = 6^3\\ 73 = 1 + 8 + 64 &\Longrightarrow 1 \times 8 \times 64 = 8^3\\ \end{align} $$ 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.

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.

Btw Im curious about this Ghandi person though info about that does not get the bounty naturally.

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.

  • 0
    What to you mean by "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"? 11 = 4 + 5 + 22012-06-15
  • 1
    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.2012-06-15
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    @MattPressland! you are absolutely true. other than 2, 5 and 11, remaining primes up to 9500, can be written as sum of three integers and the product of these numbers is cube of some integer. How to generalize or how to check this by computer based algorithm .2012-06-15
  • 0
    If n is a prime. can we write this n as sum of m numbers and product of these m numbers is sum cube? if m is some three numbers individually then n = $(product of these all members)^3$. If n = a + b+ c+ d+ e+ f then n = $(abcdef)^6$. Like this can we generalize? can we give proof mathematically or computationally? Please...answer...2012-06-15
  • 1
    The natural generalization is as follows. Let $k$ be a positive integer. Then there is an integer $N_k$ such that if $p$ is any prime greater than $N_k$, then there exist positive integers $x_1,x_2,\dots,x_k$ such that $x_1+x_2+\cdots+x_k=p$ and $x_1x_2\cdots x_k$ is a perfect $k$-th power. Like many assertions that mix primes with addition, this may be very difficult.2012-06-18
  • 2
    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.2012-06-18
  • 0
    If we let $k>m$ in @André's second generalization eventually a solution to Waring's problem suffices. And for the first generalization if there's a solution for prime $p$ then there's trivially a solution for $cp$ by multiplying each $x_i$ by $c$.2012-06-18
  • 0
    Here is an observation. Let $A_{p}=\{(x,y,z) \in \mathbb{N}^{3}: \, x+y+z=p\}$. Then $|A_{p}|=\binom{p-1}{2}=\frac{1}{2}(p-1)(p-2)$ by Stars and Bars. Let $f(x,y,z)=xyz$. Subject to the constraint that $x+y+z=p$ and $(x,y,z) \in \mathbb{N}^{3}$, the minimum of $f$ occurs at $x=y=1$ and $z=p-2$, so $\min(f)=p-2 \approx p = \left(p^{\frac{1}{3}}\right)^{3}$. The maximum occurs at $x=y=z=\frac{p}{3}$, so $\max(f)=\left(\frac{p}{3}\right)^{3}$. The number of integers that are cubes and lie between $p$ and $\left(\frac{p}{3}\right)^{3}$ is roughly $\frac{p}{3}-p^{\frac{1}{3}}$.2012-06-18
  • 0
    ...with so many tuples to choose from, it is unlikely that the intersection is empty, so I'd expect the OP's conjecture to be true and probably difficult to prove, as it seems to be the case with many problems in prime number theory.2012-06-18
  • 2
    For the original question we can choose $\{x^2,2xy,4y^2\}$ then the sum is $(x+y)^2+3y^2$ which gets a lot of (most?) primes $p=6k+1$. Or $\{x^2,4xy,2y^2\}$ then the sum is $(x+2y)-2y^2$ which gets most (all?) primes $p=8k\pm 1$. In general $\{x^2,2b^2xy,4by^2\}$, or $\{x^2,4b^2xy,2by^2\}$ or $\{x^2,2^cxy,2^dy^2\}$ give variants of $x^2\pm Ny^2$ which I thought might lead to a solution, but there are gaps.2012-06-18
  • 0
    @JacksonWalters I think the analysis is a little bit more delicate than you're giving it credit for - I was thinking along similar lines and writing up a heuristic analysis, but the precise calculations get (un)surprisingly tricky.2012-06-18
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    @JacksonWalters For instance, note that you're (naively) throwing $\Theta(p^2)$ darts at $\Theta(p)$ things over an interval of size $\Theta(p^3)$; since any given dart then hits with (naive) probability $\Theta(p^{-2})$, the probability that no such collision exists would appear at first blush to approach a non-zero constant as $p\rightarrow\infty$.2012-06-18
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    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*.2012-06-18
  • 1
    I should add that, for computational testing, it makes sense to only check primes. But there is no reason to think that this will be a useful idea in the proof.2012-06-18
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    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 in2012-06-18
  • 1
    ..either $(3\mathbb{Z}+1) \times (3\mathbb{Z}+2)$, $(3\mathbb{Z}+2) \times (3\mathbb{Z}+1)$, or $3\mathbb{Z} \times 3\mathbb{Z}$.2012-06-18
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    Every prime that is $1 \mod 3$ is of the form $a^2+ab+b^2$, so that covers half the primes immediately.2012-06-18
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    @DavidSpeyer That is a very nice observation!2012-06-18
  • 0
    What happens when k=2?2012-06-25

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