nt.number theory - A prime sequence can be partitioned into two sets of equal or consecutive sum

Scott Carnahan had an interesting idea; let's formalize it into an actual solution. We will show that, given $n ge 2$ a positive integer, $p_1, cdots, p_n$ the first n primes, we have some $e_1, cdots, e_n$ with $e_i = pm 1$ such that $|e_1p_1 + e_2p_2 + cdots + e_np_n| le 1$. (Note that we may further stipulate that $e_n = 1$.) A simple parity argument from here suffices to prove the conjecture.

We will prove this by induction on $n$. The cases $2 le n le 6$ are trivial to verify, and were provided already by a-boy. We now fix $n ge 7$.

We first need some asymptotics in the form of the Bertrand-Chebyshev theorem; we use the formulation that for $m > 1$ there is a prime between $m$ and $2m$.

Write $S_k = e_np_n + e_{n-1}p_{n-1} + cdots + e_{n-k+1}p_{n-k+1}$, and let $M(k)$ be the minimum of $|S_k|$ over all tuples $(e_n, e_{n-1}, cdots, e_{n-k+1})$. We stipulated earlier that $e_n = 1$, so we have $M(1) = p_n$. Two facts that will be useful to us in the future are that (1) $M(k+1) le |M(k)-p_{n-k}|$ and (2) if $|a| le |b|$, then $min{{|a+b|,|a-b|}} le |b|$.

We claim that $M(k) le p_{n-k+1}$ for $k = 1, 2, cdots, n-2$. We prove this by induction on $k$. The claim for $k = 1$ is trivial. Now if $M(k) le p_{n-k}$, then we are done, as $M(k+1) le min{{|M(k)+p_{n-k}|,|M(k)-p_{n-k}|}} le p_{n-k}$ by fact (2).

Now suppose $p_{n-k} < M(k) le p_{n-k+1}$. Write $2m+1 = p_{n-k+1} ge p_3 = 5$, so that $m > 1$. In this case we know that $m < p_{n-k} < M(k) le 2m$. But then $M(k+1) le M(k) - p_{n-k} le 2m-(m+1) = (m-1) < p_{n-k}$ as desired.

The fact that $M(k) le p_{n-k+1}$ is eminently useful.

Indeed, we may use it to dispatch of the even case immediately. Set $k = n-6$. Then we have $M(n-6) le 17$. As all sums considered in $M(n-6)$ are sums of an even number of odd terms, we in fact have $M(n-6) le 16$ and even. Now we simply note that all odd numbers between -15 and 15 are realizable as sums and differences of the first 6 primes, which is left as an easy computational exercise.

In the odd case, we consider $k = n-5$. Then $M(n-5) le 13$. For the same parity reasons as above, we have in fact $M(n-5) le 12$. And again, we note that all even numbers between -12 and 12 are realizable as sums and differences of the first 5 primes - another easy computational exercise.

The limits of $n-6$ and $n-5$ are the best possible for our small-case analysis.

If we were to establish an algorithm for this, we could just do the greedy algorithm on choosing $e_n$, then $e_{n-1}$, and so on, each time choosing $e_k$ so as to minimize $S_{k+1}$ (or randomly if $S_k = 0$). Our claim that $M(k) le p_{n-k}$ will continue to be satisfied by the greedy algorithm, as the proof of the claim does not involve changing prior $e_i$. Thus our greedy-algorithm mimium modulus must satisfy the same inequality, and we continue until we are at $n-6$ or $n-5$, then finish as in our nonconstructive proof.