I noticed that, $$\begin{align}3(2^3 - 3^3 + 5^3) + 6 &= 18^2 \\ \text{and } \qquad 5(2^3 - 3^3 + 5^3 - 7^3 + 11^3) + 6 &= 74^2.\end{align}$$ These equations are of the form, $$6 + (2k+1)\sum_{n=1}^{2k+1}p_n^{ \ \ 3}(-1)^{n+1} = x^2$$ such that $k\in\mathbb{N}$ and $p_n$ denotes the $n^{\text{th}}$ prime number. Here, $x\in\mathbb{Z}$ by letting $k = 1$ and $k=2$, but $k = 3$ does not hold, i.e. $x\notin\mathbb{Z}$. In fact, it appears as if $k > 2$ does not hold after testing for all $k \leqslant 61.$

Could anyone prove/disprove that there exists another equation like this for some pair $(k, x)$? Does somebody have a computer that could test for values of $k > 61$, because I only use my computer's calculator. (I used to have a program until I updated my computer software and the program lacked support for it.)

Thank you in advance.


The general equation has been tested for $k < 10,000$ and the only pairs $(k, x) \in\mathbb{N}^2$ are $(1, 18)$ and $(2, 74)$. It is now a conjecture that these are the only solution pairs over the natural numbers.

Nonetheless, I did mention in the foregoing that $x \in \mathbb{Z}$, but since we are finding $x^2$ then it is also allowable for $x$ to be natural, i.e. $x\in\mathbb{N}$.

  • 1
    $\begingroup$ It seems that you forgot the $+6$ in the lhs of the general equation. I checked up to 150 without more success. $\endgroup$ – Claude Leibovici Feb 16 '18 at 11:03
  • $\begingroup$ @ClaudeLeibovici oh woops! haha sorry about that. Thank you for pointing that out :) $\endgroup$ – Mr Pie Feb 16 '18 at 11:04
  • 2
    $\begingroup$ Unless I have made a mistake in my script it seems that for $k<10000$ the only solutions $k=1,2$. What's interesting is that we basically never get close to a square, too, at least in the range I specified. P.S. For $k=2$ the sum is $74^2$, not $78^2$ $\endgroup$ – Stefan4024 Feb 16 '18 at 12:37
  • $\begingroup$ @Stefan4024 thank you for that. I thought it was $78$, last time I checked. When I put in the value again, I realised I made a typo. $\endgroup$ – Mr Pie Feb 16 '18 at 12:50

$x_3$ must satisify following equation. $${x_3}^2=74^2+(2k-4)(2^3-3^3+5^3-7^3+11^3)+(2k+1)\sum_{n=6}^{2k+1} p_n^3(-1)^{n+1}$$ $$ =74^2+4(k-2)*547+(2k+1)(2^3-3^3+5^3+・・・+p_{2k+1}^3) $$

At least since last value of this equation is multiple of 4, twin prime don't become solution.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.