Signed dilation designs cancelling K levels need K+1 primes

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_proof: Each vertex needs a fiber-mate at every cancelled level; the mates are distinct, and admissibility forces their primes to be distinct.

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Statement

Let $K\ge1$ and let $V$ be a nonempty finite set. For each $v\in V$ let $p_v>K$ be a prime, $s_v$ an integer and $c_v$ a nonzero integer, the pairs $(p_v,s_v)$ being distinct. Suppose that for every $h\in\{1,\dots,K\}$ and every integer $m$,

$$\sum_{v:\ p_vh-s_v=m}c_v=0,$$

and that some integer $n$ satisfies $n\equiv s_v\pmod{p_v}$ for every $v\in V$. Then the set $\{p_v:v\in V\}$ has at least $K+1$ elements, and the product of its elements exceeds $K^{K+1}$.

Significance

For admissible $n$ the numbers $n+p_vh-s_v=p_v(N_v+h)$, with $N_v=(n-s_v)/p_v$, are the arguments at level $h$ of a signed combination of prime dilations of shift tails, as in the alternating cube of the prime dilation page. The hypothesis says that levels $1,\dots,K$ cancel for every function in place of the totient. The cube does this with $2^K$ distinct primes; the claim shows that every such design needs at least $K+1$. When the design is averaged over a residue class with at least two representatives inside $[x/2,x]$, the product is at most $x$, which forces $K<(1+o(1))\ln x/\ln\ln x$, a vanishing fraction of the $\log_2x$ levels that determine the phase of the binary tail. The design bound page draws this consequence for Problem 249 and records its scope. The claim neither proves nor disproves the irrationality of $\sum_{n\ge1}\varphi(n)2^{-n}$. The cube is not the most economical design: the design bound page exhibits a three-level design with six distinct primes, and a four-level integer configuration with twelve points. The latter has not been realized with prime first coordinates here. The minimum number of primes for general $K$ is not determined.

Current standing and verification

The complete proof is author-recorded at tier 0, with no independent acceptance, formal verification, or novelty claim. No native L-claim is consumed, and no computation is required. The proof uses only that a prime dividing a product divides one of its factors and that distinct primes do not divide one another.