9
$\begingroup$

Let $N$ be a positive integer. Does there always exist $n>N$ such that the first $n$ primes can be divided into two sets with equal sum?

If $n$ is even, the sum of the first $n$ primes is odd, so we cannot perform the division. But if $n$ is odd, it might be possible. For instance, $2+3=5, 2+5+7=3+11, 2+3+11+13=5+7+17$.

$\endgroup$
2
$\begingroup$

Yes, this is always possible.

The meat of the proof comes from drvitek’s answer to “Can a prime sequence be partitioned into two sets of equal or consecutive sum?” on MathOverflow. I’m not going to regurgitate it here, but I’ve tried to follow it and it seems correct to me.

He proves the result that, given the first $n$ primes $\{p_1,\ldots,p_n\}$, we can find some $e_1,\ldots,e_n$ with $e_i = \pm 1$ such that $$\left\vert \sum_{i=1}^n e_i p_i \right\vert \leq 1$$ A simple parity argument tells us that if $n$ is odd, this sum must be $0$. We then partition the primes into those with coefficient $+1$ and $-1$, and lo, a partition of the first $n$ primes with equal sum.

It follows that for any $N$, we can choose any odd $n>N$ and find such a partition of the primes. (Which seems stronger than merely some $n>N$ exists.)

$\endgroup$

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.