Statement and scope

For a positive integer M define

R_M = {a in Z_{>0} : a = 0, 1, or 2 (mod M)},

with repeated residue classes identified, and define the outer skeleton

O = union over s = 3^k, k >= 0,
    R_(2s) intersect [4s^2/9, 4s^2].

Let B be a set of positive integers containing O. Suppose that for every sufficiently large power q of 3, every r in {0,1,2}, and every integer

1 <= ell <= 2q/3 - 2,

the integer

t_(q,r,ell) = (3q-1)^2 + r - 6q*ell

has a representation t_(q,r,ell) = n^2+b with b in B, n >= 0, and n = 1 or -1 (mod q). Then

limsup_(X -> infinity) B(X)/sqrt(X) >= 9/2.

Here X ranges over real cutoffs and B(X) = |B intersect [1,X]|. The hypothesis concerns only the displayed test targets. It follows, in particular, from a policy that repairs every hard transition while retaining the old root class modulo q. It allows arbitrary repair sets, dependence on q, and all eligible old roots, including the root 3q-m omitted from the earlier four-root repair model. No Jordan measurability or limiting coverage assumption is used.

This is not a lower bound for arbitrary square complements. Such a complement may use another root class for these targets, or may omit points of O. The argument does not rule out either possibility. The proof below is author-recorded, without independent acceptance.

Exact hard targets and all eligible roots

Fix q >= 9. Put M=3q-1 and L=2q/3-2. Since q is a power of 3, L is an integer. For 1 <= ell <= L,

5q^2+6q+1+r <= t_(q,r,ell) <= 9q^2-12q+1+r < M^2.

The last inequality follows from t-M^2 = r-6q*ell < 0. Also t-r = 1 (mod q). The nonnegative integers n = +/-1 (mod q) whose squares are below t are exactly

1, q-1, q+1, 2q-1, 2q+1.

Indeed t > (2q+1)^2, while the next eligible root is M=3q-1 and t<M^2. At the old scale q the bounded root is q-1; its other same-parity representative q+1 also has residual >4q^2. At every smaller power s>=3 the bounded root is s-1, and t-(s-1)^2>4s^2. The bounded root at 3q is M, with negative residual. Thus these are actual first hard transitions, rather than merely points of a relaxed parameter space.

For each of the five roots the residual b=t-n^2 is as follows. The last column is j in b = r+jq (mod 6q).

The congruences use q=3 (mod 6). All five residuals lie strictly in (q^2,9q^2). For the smallest one, at ell=L and root 2q+1,

b = q^2+2q+r > q^2.

For the largest one, root 1 and ell=1 give

b = 9q^2-12q+r < 9q^2.

Ordering the five root squares proves the same bounds for the others. The residuals from q-1 and q+1 are both >4q^2: their smallest values are 4q^2+8q+r and 4q^2+4q+r, respectively.

No candidate residual is in the outer skeleton

In the radial interval (q^2,9q^2), only the outer annuli at q and 3q can contribute. All annuli through q/3 end by 4q^2/9; all annuli from 9q onward start at least at 36q^2.

The two candidates with j=2 or 4 lie above 4q^2, outside the annulus at q. The other three have odd j; below or at 4q^2 their residues modulo 2q are r+q, which is none of 0,1,2 for q>=3. Above 4q^2, every candidate has one of the nonzero lift classes r+jq modulo 6q, which is again none of 0,1,2. Therefore every candidate is outside O.

Distinct targets require distinct added points

For q>=3 the fifteen classes

r+jq (mod 6q),    r in {0,1,2}, j in {1,2,3,4,5},

are pairwise distinct. Thus a candidate value b determines r and the root row in the table. Within a fixed row its value decreases by exactly 6q when ell increases by one. It therefore determines ell as well.

Consequently no single b can supply an allowed representation for two different displayed targets at this q. There are 3L=2q-6 targets, so the hypothesis forces

|(B minus O) intersect (q^2,9q^2)| >= 2q-6.             (1)

This is a finite injection argument. In particular, oscillating or digit-dependent choices of which candidate to retain do not evade (1) as long as the allowed root class is unchanged.

The all-prefix counting consequence

For consecutive powers q and 3q, the intervals (q^2,9q^2) are disjoint. Hence the contributions forced by (1) at different scales do not share points. Summing through q=3^k, while ignoring a fixed initial segment, gives

|(B minus O) intersect [1,9q^2]| >= 3q-O(k).            (2)

For completeness, the outer skeleton has

O(9q^2) = (21/2)q+O(k).                               (3)

To verify (3), its annulus at a large scale s has length 32s^2/9 and three residue classes modulo 2s, hence count (16/3)s+O(1). Annuli meet only at endpoints. Summing through s=q gives O(4q^2)=8q+O(k). The additional interval (4q^2,9q^2] belongs to the annulus at 3q, with density 3/(6q), so it contributes (5/2)q+O(1). These periodic counts are uniform in the indicated endpoints.

Since the points in (2) lie outside O, addition of (2) and (3) yields

B(9q^2) >= (27/2)q-O(k).

Dividing by sqrt(9q^2)=3q and letting k tend to infinity proves the stated limsup lower bound. This is a subsequence lower bound and therefore applies directly to the limsup over all real cutoffs. It does not assert a lower bound at every cutoff.

A quantitative escape requirement

There is also a necessary condition for a general set B, without assuming that it contains O. Let D(X)=|(O minus B) intersect [1,X]|. Let E_s count the displayed targets at scale s that have no representation in B using a root congruent to +/-1 modulo s. The same finite injection gives

B(9q^2) >= (27/2)q - D(9q^2) - sum_(9<=s<=q) E_s - O(log q),

where the sum is over powers of 3. If C=limsup B(X)/sqrt(X)<9/2, then

liminf_(q=3^k -> infinity)
  [D(9q^2)+sum_(9<=s<=q) E_s]/q >= 3*(9/2-C).

Thus a proposed improvement must quantitatively remove outer-skeleton mass, use other root classes on these targets, or combine both actions. This is a necessary counting condition; satisfying it is not a coverage proof. In particular, the other root classes can come from smaller scales.

Relation to the existing frontier

The prior one-step mass calculation assumed a radial inclusion derived from m/q approaching 1. Its discrete-to-radial passage and its global prefix consequence were left unproved. The explicit targets above avoid both passages. They also include every old root below sqrt(t), so adding the previously omitted root 3q-m to that model does not remove this obstruction.

The existing L3 construction attains upper limit 9/2 while containing O and using this old-root repair policy. Its attainment argument remains the separate author-recorded L3 proof. The new lower-bound proof does not consume L3 or any external theorem.

The useful escape condition is concrete: a construction below 9/2 must either alter the outer skeleton or represent at least some of these forced targets through roots outside +/-1 modulo their corresponding q. This statement is necessary within the displayed setup, not sufficient for constructing an improvement.