bibkey: banksgaraevheathbrownshparlinski2008density authors: W. D. Banks, M. Z. Garaev, D. R. Heath-Brown, and I. E. Shparlinski year: 2008 title: Density of non-residues in Burgess-type intervals and applications doi: 10.1112/blms/bdm111 url: https://arxiv.org/pdf/math/0607692v3 claim: The proof of Theorem 2.1 gives fixed reciprocal-prime weight at a Burgess-type cutoff for prime moduli; odd affine offsets retain the prime-two contribution, while actual square-part and multiplier exceptions still require their own budget. strata_touched: [] license: citation-only triage: anchor

Prime-conductor weights with the low-end exceptions retained

The article appeared in Bulletin of the London Mathematical Society 40 (2008), 88–96, DOI:10.1112/blms/bdm111. The inspected source is arXiv:math/0607692v3, dated 25 September 2007. The following bound is an explicit intermediate conclusion in §3.1, printed pp.4–5, in the proof of Theorem 2.1; the main theorem concerns the density of integer nonresidues. No independent full proof audit or numerical threshold is claimed here. The local mod-eight implication below was checked by a transient Lean application of existing integer facts; the analytic estimate and Robin application were not compiled. No new Lean declaration is delivered.

The weighted statement actually available

For every fixed $0<\epsilon \le 0.01$ and every sufficiently large prime $P$, the proof states

$$\sum _{\begin{array}{c}\ell \le R(P)\\ (\ell /P)=-1\end{array}}\frac{1}{\ell }\ge \epsilon ,R(P)=P^{1/(4\sqrt{e})+\epsilon /2},$$

where $\ell$ ranges over primes, including two. The threshold depends only on $\epsilon$, not a chosen subsequence of $P$. There is no growing lower cutoff in this statement. It applies to the Legendre character of prime modulus; inducing that character to a composite modulus does not make this a theorem for all composite quadratic characters.

Actual-source application and its limits

Use the FIB theory volume §§181,202 and the same-integer Euler budget. Let the primitive quadratic character associated with the actual nonsquare $D$ have odd prime conductor $P$. It is then the Legendre character ${\chi }_P(\ell )=(\ell /P)$. On odd primes not dividing $D$, this agrees with the actual symbol $(D/\ell )$. A positive example is $D=P{t}^2$ with $P\equiv 1(\mathrm{mod}4)$. Positive $D=P\equiv 3(\mathrm{mod}4)$ instead has conductor $4P$ and does not meet this prime-conductor hypothesis.

For §202’s actual affine $n=h+gU$ and $Y=\log n$, require $R(P)\le Y$. Every odd supplied prime not dividing $D$ is excluded from $n$. Set

$$E_D(P)=\sum _{\begin{array}{c}\ell \le R(P)\\ {\chi }_P(\ell )=-1\\ \ell \mid 2D\end{array}}\frac{1}{\ell }.$$

Then the conservative paper-level interface is $J_{\mathrm{miss}}(n;Y)\ge \epsilon -E_D(P)$. Square-part primes absent from the primitive conductor remain in this cost. For §181’s primitive multiplicative source $n=Cu$, every odd negative-character prime is excluded from $u$, including primes in the square part: an odd prime dividing $D$ cannot divide the primitive $u$. Its corresponding cost is the same sum with $\ell \mid 2C$, rather than an independently charged ramification factor.

The upper cutoff can meet $Y$ when $P\le (\log n{)}^{4\sqrt{e}-\delta }$ for fixed $0<\delta <4\sqrt{e}$, after choosing fixed $\epsilon >0$ small enough that $(4\sqrt{e}-\delta )(1/(4\sqrt{e})+\epsilon /2)<1$. This is a cutoff range, not an automatic Robin-safe family: the exception-adjusted weight must still pay $M(Y)+T(Y)-D_{\mathrm{pow}}(n)$ for that same $n$.

In particular, if ${\chi }_P(2)=-1$, deleting two already costs $1/2$, larger than the guaranteed $\epsilon$. A fixed exceptional set does not lose its reciprocal weight merely because $P\to \infty$. Proving that an exceptional prime actually misses $n$ can restore its contribution, but that is an additional arithmetic certificate.

Odd affine offsets certify the prime-two contribution

For the same actual affine source, assume additionally that $h$ is odd. Then $D$ is odd and $D\equiv 1(\mathrm{mod}4)$. If $n$ is even, $8\mid n(n-2h)$: writing $n=2k$, the product is $4k(k-h)$, and one of $k,k-h$ is even. The existing identity gives $(gc{)}^2\equiv D(\mathrm{mod}8)$. Since $D$ is odd, $gc$ is odd, so its square is one modulo eight. Hence

$$2\mid n⟹D\equiv 1(\mathrm{mod}8).$$

This is a local application of the existing identity and ordinary parity and odd-square facts, not a new character estimate. The standard supplementary law at two implies that a negative quadratic character value at two forces $2\nmid n$ here. This concerns the primitive character; the character induced to $4|D|$ in §202 still has value zero at two.

With odd prime conductor $P$, write $D=d{t}^2$ with $t\ge 1$ odd and

$$d=\{\begin{array}{ll}P,& P\equiv 1(\mathrm{mod}4),\\ -P,& P\equiv 3(\mathrm{mod}4).\end{array}$$

The odd square $t^2$ preserves the mod-eight class. Thus ${\chi }_P(2)=(D/2)$, and a supplied negative prime two is actually missing from $n$. Every odd supplied prime not dividing $t$ is also missing by §202. The only possible absorbed negative primes belong to the same integer’s square-part intersection. Define

$$E_{\mathrm{sq}}(P;n,t)=\sum _{\begin{array}{c}\ell \le R(P)\\ {\chi }_P(\ell )=-1\\ \ell \mid \gcd (t,n)\end{array}}\frac{1}{\ell }.$$

Under the original Banks threshold and $R(P)\le Y=\log n$, the refined paper-level interface is

$$J_{\mathrm{miss}}(n;Y)\ge \epsilon -E_{\mathrm{sq}}(P;n,t).$$

In particular $\gcd (t,n)=1$ retains the entire guaranteed weight, without requiring $t=1$. When the actual signed $D=\pm P$, the square part is one and the cost vanishes. Along such actual sources with odd $h$, $n\to \infty$, $P\to \infty$ and $P\le (\log n{)}^{4\sqrt{e}-\delta }$ for a fixed $0<\delta <4\sqrt{e}$, choose fixed $0<\epsilon \le 0.01$ satisfying the cutoff condition above. The existing Euler budget then gives the paper-level consequence

$$\mathrm{limsup}\frac{Z(n)}{e^{\gamma }\log \log n}\le e^{-\epsilon }<1.$$

This is a conditional source-family application, including negative signed $D=-P$ with $P\equiv 3(\mathrm{mod}4)$. It neither proves that all candidates meet the hypotheses nor proves that such growing-prime canonical families exist. Banks supplies no numerical onset threshold here.

The square-part condition has a real canonical obstruction. The legal unit-bit-one representation

$$n=6921=6765+144+8+3+1$$

has composition $(a,b)=(1009,1634)$, $g=h=1$, and $Q=-3169$, $D=12681=1409\cdot 3^2$. The prime conductor is $1409$, ${\chi }_{1409}(3)=-1$, and $3\mid n$. Thus negative primitive-character primes in the original square part cannot all be declared missing, even on this canonical slice. Removing oddness of $h$ also fails in the general affine model: $h=0$, $g=1$, $(a,b)=(3,2)$ give $n=12$ and $D=-44=-11\cdot 2^2$, but ${\chi }_{11}(2)=-1$ and $2\mid n$.

The Bourgain–Lindenstrauss window has a different advantage: its lower cutoff eventually exceeds two and its original $D$ retains square-part zeros. The Pollack input covers arbitrary nonprincipal characters. Its Theorem 1.1 states a prime count; the linked note also records a weighted induced-character consequence of its published inputs, with the enlarged modulus and its zeros retained. Reuse these distinct existing statements according to their actual hypotheses; none of these interfaces settles all Robin candidates or RH.