# Considering the equation, $6 + (2k+1)\sum_{n=1}^{2k+1}p_n^{ \ \ 3}(-1)^{n+1} = x^2$.

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.

Update:

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. $x\in\mathbb{N}$.

• It seems that you forgot the $+6$ in the lhs of the general equation. I checked up to 150 without more success. – Claude Leibovici Feb 16 '18 at 11:03
• @ClaudeLeibovici oh woops! haha sorry about that. Thank you for pointing that out :) – Mr Pie Feb 16 '18 at 11:04
• 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 $78^2$ – Stefan4024 Feb 16 '18 at 12:37
• @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. – Mr Pie Feb 16 '18 at 12:50

## 1 Answer

$x_3$ 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.