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We prove L3 for its displayed set $A$. Throughout, a power $q$ means $3^k$ with integer $k\ge0$. Membership in $R_M$ means positive-integer membership in one of the residue classes $0,1,2$ modulo $M$, with repetitions removed. Counting functions use the closed real cutoff convention $A(X)=|A\cap[1,X]|$.

Bounded square roots

Fix an integer $u\equiv1\pmod3$. For each power $q$, there is a root $m_q$ with

$$0\le m_q\le q,\qquad m_q^2\equiv u\pmod{2q}. \tag{6}$$

At adjacent powers $q$ and $Q=3q$, the bounded roots satisfy

$$m_Q\in\{m_q,\ 2q-m_q,\ 2q+m_q\}. \tag{7}$$

Here are complete elementary justifications. Modulo $3$, the roots of $u$ are exactly $1$ and $-1$. Suppose $q=3^k$ with $k\ge1$, and $z$ is a root modulo $q$. Write $z^2-u=dq$. The three possible lifts of its residue class modulo $3q$ are $z+hq$ for $h=0,1,2$. Their square congruence is equivalent to

$$d+2zh\equiv0\pmod3,$$

because $3\mid q$. Since $z^2\equiv1\pmod3$, the coefficient $2z$ is nonzero modulo $3$. Exactly one $h$ solves the displayed congruence. Every root modulo $3q$ reduces to a root modulo $q$. Induction therefore gives exactly two roots modulo every $q\ge3$. They are distinct sign classes: a root is not divisible by $3$, so it cannot equal its negative modulo $q$.

For any root class modulo the odd integer $q$, the two classes above it modulo $2q$ have opposite parity. Choose the one with parity $u$. Its square is then congruent to $u$ modulo both $q$ and $2$, hence modulo $2q$: if an integer is divisible by $q$ and by $2$, it is divisible by $2q$ because $q$ is odd. Conversely every root modulo $2q$ arises this way. Thus for $q\ge3$ there are exactly two sign classes modulo $2q$. Neither is represented by $0$ or $q$, since those integers are divisible by $3$. For either root representative $z\in[0,2q)$, the smaller of $z$ and $2q-z$ is the unique root $m_q$ with $0<m_q<q$.

A root $m_Q\in[0,3q]$ reduces modulo $2q$ to $m_q$ or $-m_q$. For $0<m_q<q$, all representatives of these classes in $[0,3q]$ are exactly $m_q,2q-m_q,2q+m_q$. This proves (7) when $q\ge3$.

For $q=1$, put $m_1=0$ if $u$ is even and $m_1=1$ if $u$ is odd. These are precisely the bounded roots modulo $2$. An even $u\equiv1 \pmod3$ is $4$ modulo $6$ and has bounded root $m_3=2$; an odd such $u$ is $1$ modulo $6$ and has bounded root $m_3=1$. Each belongs to the set on the right of (7). This supplies the first transition as well as the starting case of (6).

Catching a positive residual

If $a$ is an integer, $r\in\{0,1,2\}$, and

$$0<a\le4q^2,\qquad a\equiv r\pmod{2q}, \tag{8}$$

then $a\in A$. To see this, list the closed outer intervals at scales $1,3,\ldots,q$. The upper endpoint $4s^2$ at scale $s$ equals the lower endpoint $4(3s)^2/9$ at scale $3s$. Their union is $[4/9,4q^2]$, which contains every positive integer at most $4q^2$. Choose a scale $s\le q$ whose interval contains $a$. Since $2s$ divides $2q$, congruence (8) implies $a\equiv r\pmod{2s}$. The outer annulus at $s$ contains $a$. This argument includes shared endpoints and $s=1$.

Coverage of every positive target

At scale $q=1$, $R_2$ contains every positive integer and the outer interval contains $1,2,3,4$. Thus these four targets have the form $0^2+a$, and $5=1^2+4$.

Fix an integer $t\ge6$. Choose the unique $r\in\{0,1,2\}$ with $r\equiv t-1\pmod3$, and put $u=t-r$. Apply (6) and (7) to this $u$, and write

$$a_q=t-m_q^2. \tag{9}$$

Each $a_q$ is an integer congruent to $r$ modulo $2q$. If any $a_q$ lies in $[4q^2/9,4q^2]$, it belongs to the outer annulus at $q$ and represents $t$. We may therefore assume that no $a_q$ lies in that closed interval, and produce another representation.

Since $m_1\in\{0,1\}$, we have $a_1\ge5>4$. Also $a_s\le t$ for every scale $s$. There is consequently a least power $Q>1$ such that $a_Q\le4Q^2$. Put $Q=3q$. Minimality gives $a_q>4q^2$. At scale $Q$ the residual is at most the outer upper endpoint, and by assumption it is outside the closed outer interval. Hence it lies strictly below the lower endpoint:

$$a_q>4q^2,\qquad a_Q<\frac{4Q^2}{9}=4q^2. \tag{10}$$

If $a_Q>0$, it satisfies (8) at scale $q$, because its congruence modulo $2Q$ descends modulo $2q$. It then belongs to $A$ and represents $t$. We may thus assume $a_Q\le0$, including the possibility $a_Q=0$.

Write $m=m_q$ and $M=m_Q$. Formula (7) lists all possibilities for $M$. The choice $M=m$ contradicts (10). The choice $M=2q-m$ would imply

$$a_Q=a_q-4q^2+4qm>0,$$

contrary to $a_Q\le0$. Therefore $M=2q+m$. Define the alternative residual

$$e=t-(2q-m)^2=a_q-4q^2+4qm>0. \tag{11}$$

The integer $2q-m$ is a root of $u$ modulo $2q$, so $e\equiv r\pmod{2q}$. If $e\le4q^2$, (8) gives $e\in A$ and $t=(2q-m)^2+e$. It remains to treat $e>4q^2$.

Set

$$b=t-(q-m)^2=a_q+2qm-q^2. \tag{12}$$

Because $q-m=Q-M$ and $M^2\equiv u\pmod Q$, we have $b\equiv r\pmod Q$. Moreover (10) and (11) give

$$b=\frac{a_q+e}{2}+q^2>5q^2. \tag{13}$$

The inequality $a_Q=a_q-4q^2-4qm\le0$ gives the upper bound

$$b\le3q^2+6qm\le9q^2, \tag{14}$$

where the last step uses $m\le q$. Thus

$$b\in R_Q\cap\left[\frac{5Q^2}{9},Q^2\right]\subseteq A.$$

Finally $q-m\ge0$, so (12) is a representation by a nonnegative square and an element of $A$. The argument allowed equality in (14), and allowed $a_Q=0$ throughout its final branch. Together with the five initial targets, it proves positive-target coverage.

Exact cycles and their endpoint conventions

For each power $q$, let

$$C_q=(L_q,U_q]\cap\mathbb Z_{>0},\qquad L_q=\frac{4q^2}{9},\qquad U_q=4q^2. \tag{15}$$

The cycles are pairwise disjoint and partition the positive integers: $U_q=L_{3q}$, their union as real intervals is $(4/9,\infty)$, and every positive integer exceeds $4/9$. The first cycle is $C_1=\{1,2,3,4\}$, all of which belong to $A$.

Fix $q\ge3$. An outer annulus from a smaller scale has upper endpoint at most $L_q$ and cannot meet $C_q$. An outer annulus from a larger scale has lower endpoint at least $U_q$; only the next scale can meet $C_q$, and only at $U_q$. That endpoint already lies in the outer annulus at $q$, because $U_q=4q^2\equiv0\pmod{2q}$. It introduces no extra member. Every inner annulus from a smaller scale ends at most at $q^2/9<L_q$. Every inner annulus from a larger scale starts at least at $5q^2>U_q$. Thus on $C_q$ the literal union defining $A$ has exactly the membership given by its two annuli at scale $q$.

Put $B_q=5q^2/9$ and $D_q=q^2$. These are integers for $q\ge3$. Since $R_{2q}\subseteq R_q$, we obtain the following disjoint description, including all internal endpoints:

$$\begin{aligned} A\cap C_q={}& \bigl(R_{2q}\cap(L_q,B_q)\bigr)\\ &\mathbin{\dot\cup} \bigl(R_q\cap[B_q,D_q]\bigr)\\ &\mathbin{\dot\cup} \bigl(R_{2q}\cap(D_q,U_q]\bigr). \end{aligned} \tag{16}$$

The point $L_q$ belongs to the preceding cycle and is deliberately excluded from (16). For instance, at $q=3$ it is $4\in A$, although $4\notin R_6$. Its contribution is retained in the prefix count below. For integer endpoints $L<U$, the corresponding half-open prefix identity would be

$$A\cap[0,U) =(A\cap[0,L))\mathbin{\dot\cup}(A\cap\{L\}) \mathbin{\dot\cup}(A\cap(L,U-1]).$$

Replacing its final piece by $A\cap(L,U]$ changes the right endpoint. Below we use closed prefixes throughout, so the exact new piece is $A\cap(L_q,X]$, and it is the full cycle only when $X=U_q$.

Periodic counts with a uniform error

For a residue $r$ modulo an integer $M\ge1$ and real $a\le b$, the number of integers congruent to $r$ in $(a,b]$ is exactly

$$\left\lfloor\frac{b-r}{M}\right\rfloor -\left\lfloor\frac{a-r}{M}\right\rfloor. \tag{17}$$

It differs from $(b-a)/M$ by at most $1$. When $M\ge3$, the three residues defining $R_M$ are distinct. Summing (17) shows that their count in a positive interval $(a,b]$ differs from $3(b-a)/M$ by at most $3$. Changing either endpoint's inclusion alters the count by at most one. Consequently, for any choices of open or closed endpoints in a positive interval of length $\ell$, the count from $R_M$ differs from $3\ell/M$ by at most $6$. This also covers a singleton interval, an empty interval, and an interval whose endpoints coincide. This bound does not depend on $M$, the endpoint values, or the translation of the interval.

For $q\ge3$, both moduli in (16) are at least $3$. Apply this estimate to the three intervals there. Their lengths are $q^2/9$, $4q^2/9$, and $3q^2$, respectively. Therefore

$$|A\cap C_q|=6q+\varepsilon_q,\qquad |\varepsilon_q|\le18, \tag{18}$$

since

$$\frac{3}{2q}\frac{q^2}{9} +\frac3q\frac{4q^2}{9} +\frac{3}{2q}(3q^2)=6q.$$

Now let $q=3^k$ with $k\ge1$. The cycles preceding $C_q$ are $C_1,C_3,\ldots,C_{q/3}$ and partition all positive integers at most $L_q$. The exact first-cycle count is four. Summing (18) for the remaining $k-1$ preceding cycles gives

$$\begin{aligned} A(L_q) &=4+\sum_{j=1}^{k-1}(6\cdot3^j+\varepsilon_{3^j})\\ &=3q-5+\sum_{j=1}^{k-1}\varepsilon_{3^j}. \end{aligned} \tag{19}$$

The sum is empty when $k=1$. In particular,

$$|A(L_q)-3q|\le5+18(k-1). \tag{20}$$

For an arbitrary real $X\in[L_q,U_q]$, intersect each interval in (16) with $(-\infty,X]$. The resulting lengths are

$$\begin{aligned} \ell_1(X)&=\max\{0,\min\{X,B_q\}-L_q\},\\ \ell_2(X)&=\max\{0,\min\{X,D_q\}-B_q\},\\ \ell_3(X)&=\max\{0,X-D_q\}. \end{aligned}$$

Each nonempty intersection is an interval with its actual endpoint conventions. Applying the same bound, including degenerate cases at the breakpoints, proves

$$\left| |A\cap(L_q,X]|-T_q(X) \right|\le18, \qquad T_q(X)=\frac{3\ell_1(X)}{2q} +\frac{3\ell_2(X)}q+\frac{3\ell_3(X)}{2q}. \tag{21}$$

The equality $A(X)=A(L_q)+|A\cap(L_q,X]|$ is exact. Combining (20) and (21) yields

$$|A(X)-(3q+T_q(X))|\le18k+5. \tag{22}$$

On putting $x=X/q^2$, direct substitution of the three lengths gives $3q+T_q(X)=qF(x)$, with $F$ as in the statement. For example, its three normalized expressions are

$$3+\frac32\left(x-\frac49\right),\qquad 3+\frac16+3\left(x-\frac59\right),\qquad 3+\frac16+\frac43+\frac32(x-1).$$

They simplify to the three formulas for $F$. This proves the asserted uniform estimate, including both ends of every cycle. It also explains why using a count only at the cycle endpoints would be insufficient.

Both exact extremal limits

Set $H(x)=F(x)/\sqrt x$ for $4/9\le x\le4$. For every real cutoff in (22), $\sqrt X=q\sqrt x\ge2q/3$, so

$$\left|\frac{A(X)}{\sqrt X}-H(X/q^2)\right| \le\frac{3(18k+5)}{2\cdot3^k}. \tag{23}$$

The right-hand side tends to zero as $k\to\infty$. For example, $k\le2^k$ for $k\ge1$, by induction, and hence $k/3^k\le(2/3)^k\to0$. Every real $X\ge4$ lies in one of the intervals $[4q^2/9,4q^2)$ with $q=3^k$, $k\ge1$, and the selected $k$ tends to infinity with $X$. Thus (23) is an error tending to zero uniformly over all sufficiently large real cutoffs, including seams.

To locate the extrema, for positive constants $a,b$ differentiate

$$\frac{a}{\sqrt x}+b\sqrt x: \qquad \frac{d}{dx}\left(\frac{a}{\sqrt x}+b\sqrt x\right) =\frac{bx-a}{2x^{3/2}}.$$

On the first interval $[4/9,5/9]$, the values $a=7/3$, $b=3/2$ make this derivative negative. The profile decreases from $9/2$ to $19/(2\sqrt5)$. On $[5/9,1]$, the values $a=3/2$, $b=3$ make the derivative positive, so the profile increases from $19/(2\sqrt5)$ to $9/2$. The common minimum of these two pieces exceeds $3\sqrt2$, because both are positive and

$$\left(\frac{19}{2\sqrt5}\right)^2 =\frac{361}{20}>18=(3\sqrt2)^2.$$

On $[1,4]$, the profile is $3/\sqrt x+(3/2)\sqrt x$. Its derivative is negative for $x<2$, zero at $2$, and positive for $x>2$. It has value $9/2$ at both endpoints and minimum $3\sqrt2$ at $x=2$. Consequently

$$3\sqrt2\le H(x)\le\frac92 \quad\left(\frac49\le x\le4\right). \tag{24}$$

The uniform error (23) and (24) give

$$\liminf_{X\to\infty}\frac{A(X)}{\sqrt X}\ge3\sqrt2, \qquad \limsup_{X\to\infty}\frac{A(X)}{\sqrt X}\le\frac92.$$

For $X_k=q^2=3^{2k}$ the normalized argument is $1$ and $H(1)=9/2$. Equation (23) therefore shows $A(X_k)/\sqrt{X_k}\to9/2$. For $Y_k=2q^2=2\cdot3^{2k}$ it shows $A(Y_k)/\sqrt{Y_k}\to3\sqrt2$, because $H(2)=3\sqrt2$. Both sequences tend to infinity. They supply the reverse inequalities for the upper and lower limits, proving both equalities in the statement.

Dependencies

All essential deductions are given above using elementary integer and real arithmetic. The argument uses no external theorem, no native L-claim, and no computational premise. Its current verification standing is recorded on the claim page.