Rough shadow energy

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A different attempted proof attached a rough-quotient shadow to each private endpoint and sought an averaged packing estimate. The earlier obstruction rules out bounded pointwise overlap. The construction below also rules out a uniform second-moment substitute, even with globally private positive witnesses and bounded total shadow measure.

Statement

There are finite distinct-modulus singleton families with globally private endpoints $r_q\in[q,2q)$ for which the rough shadows

$$S_q=\{a_q+qt:t\ge1,\ P^-(t)>q\},\qquad P^-(1)=\infty,$$

obey

$$\sum_q d(S_q)\le C, \qquad \int\left(\sum_q\mathbf1_{S_q}\right)^2\,d\mu\longrightarrow\infty. \tag{1}$$

In the integral, each shadow denotes its full periodic congruence version; its natural density and pair intersection densities are unchanged by discarding the finite inactive part. Thus a uniform bound $\int Z^2\le C\int Z$, or even a uniform bound on $\int Z^2$ when $\int Z$ is bounded, is false for this proposed shadow assignment.

A private fixed-rank family

Let $\mathcal P$ be the odd primes at most $y$, and put $\lambda=\sum_{p\in\mathcal P}1/p$. For an integer $k\ge1$, let $D_k$ be all products of exactly $k$ distinct primes of $\mathcal P$. Use $a_q=q-1$ and $r_q=2q-1$.

Every $q$ is odd, and $D_k$ is primitive. If $r_p$ belongs to the $q$-cylinder, then $q\mid2p$, hence $q\mid p$. Equal squarefree rank forces $p=q$. The witnesses are therefore globally private, not just private relative to earlier moduli. Activation is exact at $r_q=2q-1$.

One shadow has density

$$d(S_q)=\frac1q\prod_{p\le q}(1-1/p)\asymp\frac1{q\log q}. \tag{2}$$

The constants follow from the elementary Euler-product bounds.

Uniformly in $k$ and $y$,

$$\sum_{q\in D_k}\frac1{q\log q}\le3. \tag{3}$$

Indeed, write $A(t)=\sum_{p\ge3}p^{-1-t}$ for $t>0$ and use $1/(q\log q)=\int_0^\infty q^{-1-t}\,dt$. The elementary symmetric sum is at most $A(t)^k/k!$. For $0<t\le1$,

$$A(t)\le\log\zeta(1+t)\le\log(1+1/t)\le\log(1/t)+\log2.$$

Substitution $u=\log(1/t)$ bounds its integral by

$$\frac1{k!}\int_0^\infty(u+\log2)^k e^{-u}\,du\le2.$$

For $t\ge1$, put $C_0=\sum_{n\ge3}n^{-2}<1$; then $A(t)\le C_0 3^{1-t}$, whose $k$th power divided by $k!$ has integral $C_0^k/(k!k\log3)<1$. This proves (3), and hence the bounded mean in (1).

Pair intersections remain large

Let $g=(q,p)$, $q=gu$, $p=gv$, and $L=\operatorname{lcm}(q,p)$. The two cylinders require $n+1=Lz$. Their quotient variables in the shadows are respectively $vz-1$ and $uz-1$. At the prime $2$, both leading coefficients are odd, so the two forbidden roots coincide. At an odd prime at most $\min(q,p)$, at most two roots are forbidden; at primes between the two cutoffs, at most one is forbidden. A zero leading coefficient gives the constant $-1$, and excludes nothing. CRT therefore yields

$$\mu(S_q\cap S_p) \ge\frac{c}{L\log q\log p}. \tag{4}$$

For detail, the conditional probability given $n+1\in L\widehat{\mathbb Z}$ is bounded below by

$$\frac12\prod_{3\le\ell\le\min(q,p)}(1-2/\ell) \prod_{\min(q,p)<\ell\le\max(q,p)}(1-1/\ell),$$

where the products are over primes. The first product is bounded below by an absolute positive constant times the square of $\prod_{\ell\le\min(q,p)}(1-1/\ell)$, since the correction ratios form a convergent positive product. The displayed quantity is therefore $\gg1/(\log q\log p)$, proving (4).

Selecting a rank with large gcd energy

Over all pairs of squarefree products of primes from $\mathcal P$, the sum of $1/\operatorname{lcm}(q,p)$ is

$$Z=\prod_{\ell\in\mathcal P}(1+3/\ell) \asymp(\log y)^3.$$

Normalize these weights to a probability law. At each prime the four states (neither product, first only, second only, both) have weights $1,1/\ell,1/\ell,1/\ell$, independently between primes. If $K_1,K_2$ are the two ranks, their difference has independent symmetric increments, with probabilities $1/(\ell+3)$ each for $+1$ and $-1$. Its characteristic function is

$$\varphi(t)=\prod_{\ell\in\mathcal P} \left(1-\frac{2(1-\cos t)}{\ell+3}\right).$$

Every factor is positive for real $t$. For $|t|\le\lambda^{-1/2}$ and large $\lambda$, the product is at least an absolute positive constant. Fourier inversion consequently gives

$$\Pr(K_1=K_2)\ge c/\sqrt\lambda.$$

Also $\mathbb E K_i\le2\lambda$, and the exponential-moment bound at parameter $\log2$ gives

$$\Pr(K_i>4\lambda)\le\exp(-(4\log2-2)\lambda).$$

The probability of rank zero for both products is $1/Z$, also negligible. Pigeonholing among the at most $4\lambda$ positive ranks up to $4\lambda$ supplies one $k$ with

$$1\le k\le4\lambda, \qquad \sum_{q,p\in D_k}\frac1{\operatorname{lcm}(q,p)} \ge\frac{cZ}{\lambda^{3/2}}. \tag{5}$$

Every member of $D_k$ satisfies $\log q\le k\log y$. Equations (4)--(5) now give

$$\int\left(\sum_{q\in D_k}\mathbf1_{S_q}\right)^2d\mu \ge\frac{cZ}{\lambda^{3/2}k^2(\log y)^2} \gg\frac{\log y}{(\log\log y)^{7/2}}\longrightarrow\infty.$$

Together with (2)--(3), this proves (1).

The residues are all the fixed signed center $-1$, a solved logarithmic density regime. The construction refutes this specific quadratic shadow charge; it does not refute endpoint nullity, the primitive-type reciprocal sum target, or a different fractional allocation of shadows.