Does $\exists n : 6 + \sum_{i=2}^n p_i = x^6+y^6$ for $p_i$ the $i^{\text{th}}$ prime and $x,y\in\mathbb{Z}$?

Question

I noticed something about the prime numbers:

Pick the number $2$. Then, add the first odd prime, namely $3$. The result is $5 = 1^2+2^2$. Notice that the exponents are also the number we picked.

Pick the number $4$. Add the first odd prime, then the second, and the third, all the way up to $71$. It follows that $$4+3+5+\cdots + 67+71=5^4 + 2^4.$$

I cannot find a number $n$ such that when you add all the $n$ odd primes together, and then add $6$ (the next even number from $4$) then the result is $x^6+y^6$ for $x,y\in\mathbb{Z}$. Does there exist a number $n$ for this specific case? Is there a more mathematical way of finding out as opposed to brute-force?

Thank you in advance.

Answer 1

Let's look at some heuristics.

Pick some x and y. Then start with six, add consecutive primes, until the sum is ≥ $x^6 + y^6$. In other words, 6 + sum of odd primes up to $p_{i-1}$ is less than $x^6 + y^6$, and adding 6 + sum of odd primes up to $p_i$ is ≥ $x^6 + y^6$.

Let d = (6 + sum of odd primes up to $p_i$) - $(x^6 + y^6)$. Obviously $0 ≤ d < p_i$, and we have a solution if d = 0. Heuristically, we assume that d could be any of the numbers from 0 to $p_i - 1$ with equal probability $1 / p_i$.

Let $z = x^6 + y^6$. The i-th prime is quite roughly equal to $i · log i$, the sum of the first i primes is quite roughly $i^2 · log i / 2$, which makes $i^2 · log i ≈ 2z$, $i ≈ 2 (z / log z)^{1/2}$, $p_i ≈ (z · log z)^{1/2}$.

The chance that there is a solution for x, y is about $1 / (z · log z)^{1/2}$. Assuming x ≥ y, this is about $1 / (x^3 · (6 log x)^{1/2})$. If we add this for $0 ≤ y ≤ x$, the sum is about $1 / (x^2 · (6 log x)^{1/2})$. The sum of $1 / x^2$ is just π / 6, this sum is smaller, maybe 1/4.

A program that checks all cases with x ≤ 100 and therefore $z < 2·10^{12}$ is doable and has a chance of finding a solution (rough estimate 25%). Above that it is quite possible that a solution exists, but it would be unlikely (I'd say probability about 0.17%), and a solution would be hard to find.

If you examined all cases up to x ≤ 10,000, that would be a ridiculous amount of work (finding all primes up to $2·10^{24}$), and the chance that you still missed a solution is still about 0.0003%.

So a brute force search adding the primes up to $2·10^{12}$ has a reasonable chance of finding a solution. But even if you process all primes up to $2·10^{24}$, there is still a non-neglible chance that there is a solution further out.

Answer 2

Heuristically: there are approximately $\Theta(\sqrt[3]{n})$ numbers $\leq n$ that are the sum of two sixth powers. (This is because there are $\Theta(\sqrt[6]{n})$ sixth powers less than $n$ and we're looking at the size of the set $S_{6,2}=\{x+y: x\in S_6, y\in S_6\}$ where $S_6$ is the set of sixth powers; generically we should expect that if we define $S_2=\{x+y: x, y\in S\}$ and $S$ is sparse enough, then $|S_2|$ will have size $\Theta(|S|^2)$, and this can be made rigorous.) This means that we can think of the 'probability' that a random number of size $\approx N$ is a sum of two sixth powers as being approximately $\Theta(N^{-2/3})$.

Now, the size of the $n$th prime is approximately $n\log n$, so the sum of the first $n$ primes is roughly $\Theta(n^2\log n)$ (note that the initial term of $6$ doesn't impact these asymptotics at all). Thus, using the probability estimate above, the probability that this is a sum of two sixth powers is $\Theta\left((n^2\log n)^{-2/3}\right)$ or $\Theta\left(n^{-4/3}(\log n)^{-2/3}\right)$. Now, by linearity of expectation, the total number of solutions should be the sum of this probability over all possible $n$; in other words, $\sum_{n=1}^\infty a_n$ where $a_n\in\Theta\left(n^{-4/3}(\log n)^{-2/3}\right)$. But by the usual $p$-series comparison, we would expect this series to converge; in other words, we 'should' expect only a finite number of solutions in total. This suggests that that finite number might be zero, so it is very possible that you won't find any solutions.

(Note that running the same asymptotic heuristic in the case of fourth powers yields the probability that the $n$th prime sum is a sum of fourth powers being roughly $\Theta\left((n^2\log n)^{-1/2}\right)$ or $\Theta\left(n^{-1}(\log n)^{-1/2}\right)$, and then the corresponding sum over all $n$ diverges, so we should expect infinitely many examples there.)