# About prime numbers and the sums of three distinct cubes

I am not a professional mathematician, but I believe this question is not something unprofessional (for example, like solving for $x$ in the following: $\sqrt{x} + \sqrt{15 + x} = 15$) if you know what I mean. The reason being is because I cannot find solutions to the following equation through a not so exhausting algebraic method (for example, trial and error): $$a^3 + b^3 + c^3 = \left\{d^3 : a, b, c, d \in \mathbf{P}\right\}$$ $\mathbf{P}$ denotes the set of prime numbers, and I can only try to find solutions through some computer power using Ramanujan's Formula for certain integer solutions to the equation. This is where I found only one solution for which $0 < a < b < c \leqslant 2000$. $$193^3 + 461^3 + 631^3 = 709^3$$ However I grew curious to see whether there are other solutions. I have found many other near-solutions where $a, b, c, d$ are all prime except for one of the variables. Let the non-prime equal $k$ where $k = a, b, c$ or $d$, then it seems like $k$ is always the product of two primes, thus $k$ is semi-prime. And in my search, I have only come across one solution where $k$ is the product of two adjacent (neighbouring) primes. $$109^3 + 293^3 + 437^3 = \left\{479^3 : k = 437 = 19\times 23\right\}$$ I have only found eight-near solutions so far, but I did overlook solutions where $a, b, c, d$ are odd but $5$ divides one of them, due to a speculation that I had made.

If $5$ divides $a, b, c$ or $d$, then there exists (at least) another integer amongst $a, b, c, d$ that is (also) not prime. For example, $3^3 + 4^3 + 5^3 = 6^3$.

However, funnily enough, I found a counter-example (also a near-solution) right near the end of my search just by pure coincidence. $$415^3 + 1627^3 + 1997^3 = 2311^3$$ Because of this, I also tried looking for solutions for which $a, b, c, d$ were all odd, but none of them were prime, and I found two solutions, although there could be many more. $$147^3 + 471^3 + 1403^3 = 1421^3$$ $$255^3 + 539^3 + 1477^3 = 1503^3$$ And then during this particular search, I made two more speculations.

If $3$ divides $a, b, c$ or $d$, then $3$ also divides another integer amongst $a, b, c, d$.

Every perfect number cubed can be expressed as the sum of three cubes.

I am certain that $709$ is the smallest prime to be written as the sum of three cubes $a^3 + b^3 + c^3$ for which $a, b, c$ are prime, but that is assuming there exists other solutions. So my question is, does there exist other solutions? Is there an example of a near-solution for which $k$ is not semi-prime? Does there exist a counter-example to my two most recent speculations?

Thank you in advance.

Edit: I found another solution to $a^3 + b^3 + c^3 = d^3$ for which $a, b, c, d$ are prime. $$103^3 + 2179^3 + 2213^3 = 2767^3$$ and credit to Gerry Myerson for finding another smaller solution: $$599^3 + 691^3 + 823^3 = 1033^3$$ so there you go, and there exists a near-solution where $k$ is not semi-prime. Here, $k = 605 = 5\times 11^2$. $$277^2 + 605^2 + 2111^3 = 2129^3$$ But, in this counter-example, along with all other examples, $\Omega(k) = 2$. Perhaps I could extend the original speculation for which at every near solution, $\Omega(k) = 2$.

And, I forgot to add another speculation I have made to this post: If you have the equation, \begin{align} a^3 + b^3 + c^3 &= \left\{d^3 : \min\{a,b,c\} = 6\right\} \\ \implies 6 &\mid d + (-1)^{n + 1}\end{align} I also discovered other interesting properties with perfect numbers. \begin{align} 3^3 + 4^3 + 5^3 &= 6^3 \\ 18^3 + 19^3 + 21^3 &= 28^3 \\ 57^3 + 82^3 + 495^3 &= 496^3 \\ 2979^3 + 4005^3 + 7642^3 &= 8128^3 \end{align} $3$ divides all values of $a$, and it appears that: $$\bigg(\frac{a + b + c}{2}\bigg)^3 = \left\{x^3 + y^3 + z^3 : x, y, z \in \mathbb{N}\right\}$$ 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 $x = 162, y = 173, z = 282$.

And there must exist at least one prime amongst $x, y, z$ in each equation, but this raises another question; can $x, y, z$ all be prime?

• This question certainly isn’t “unprofessional” and I suspect most analytic number theorists would be quite interested in this problem. That said, I doubt much can be proved because this problem is just outside what can be done by techniques in diophantine geometry – Stanley Yao Xiao Oct 7 '17 at 4:33
• Here's one somewhat related interesting result. A long time back Hardy and Littlewood conjectured that infinitely many primes are the sum of three cubes. This was finally established by Heath-Brown who showed that there are infinitely many primes of the form $x^3+2y^3=x^3+y^3+y^3$. Making the variables also prime seems beyond current technology. – Lucia Oct 7 '17 at 4:55
• You will be interested in oeis.org/A258262 – note, for example, $1033^3 = 599^3 + 691^3 + 823^3$ – Gerry Myerson Oct 7 '17 at 11:49
• This is even better: oeis.org/A114923 and oeis.org/A114923/b114923.txt – Gerry Myerson Oct 7 '17 at 12:21
• @GerryMyerson Thank you Gerry for your links and information and also the smaller found solution for which $a, b, c, d$ are prime.........WOW these sites are EXACTLY what I need!! :) – user477343 Oct 7 '17 at 21:41