A biased block and an amplified moment

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

This lesson and the next three prove that the residue supports $S_p$ of Lesson 1 are not biased by quadratic characters, even when the character is translated by a different residue at every prime (Theorem 1.1). The proof is by contradiction. A bias at many primes produces, through an amplified moment and Poisson summation, a quadratic exponential sum that must be large for many products of primes. Lessons 4 and 5 turn this into a single rational number $h/m$ that serves as the translating centre at many primes simultaneously. Lesson 6 then uses the quadratic large sieve to show that the two summands concentrate on few quadratic kernels, and the collision estimate of Lesson 2 rules this out.

This lesson sets up the contradiction hypothesis, chooses the scales, proves a combinatorial identity that removes repeated primes from a moment, and computes the Poisson dual of the moment.

Prerequisites. Square-root bounds and collision stability, Sections 4 and 5; Gauss sums, Theorems 1.1 and 2.1; Poisson summation, theta, and the functional equation, Theorem 1.2; Counting primes by elementary means: Chebyshev and Mertens, Theorem 3.1.

1. The statement

For an odd prime $p$ let $\chi_p$ be the quadratic character of $\mathbb F_p$, extended by $\chi_p(0)=0$. Keep the standing assumption of Lesson 1 and the supports $S_p$.

Theorem 1.1 (quadratic decorrelation [OpenAI-013, Proposition 3.1]). As $T\to\infty$, $\begin{aligned} &\sum_{T\le\log p\le2T}\frac{\log p}p\\ &\qquad \, \max_{t\in\mathbb F_p}\Bigl|\mathbb E_{x\in S_p}\chi_p(x-t)\Bigr|\\ &\quad =o(T).\quad(1.1) \end{aligned}$

The translating residue $t$ is chosen separately at every prime. Nothing is assumed to relate the choices at different primes; relating them is the main difficulty.

Corollary 1.2 ([OpenAI-013, Corollary 3.2]). For fixed $0<\alpha<\beta$, as $L_*\to\infty$, $\begin{aligned} &\sum_{\alpha L_*\le\log\log p\le\beta L_*}\frac1p\\ &\qquad \, \max_{t\in\mathbb F_p}\Bigl|\mathbb E_{x\in S_p}\chi_p(x-t)\Bigr|\\ &\quad =o(L_*). \end{aligned}$

Proof, assuming Theorem 1.1. Cover the range of $\log p$, which is $[e^{\alpha L_*},e^{\beta L_*}]$, by $O(L_*)$ intervals $[T,2T]$ with $T\ge\frac12e^{\alpha L_*}$; positivity allows enlarging the two end intervals. On $[T,2T]$, $1/p\le(\log p)/(Tp)$, so by (1.1) the contribution of each interval is at most $\epsilon(T)$, where $\epsilon(T)\to0$. Since every $T$ used tends to infinity with $L_*$, the sum of the $O(L_*)$ contributions is $o(L_*)$. $\square$

The rest of this lesson and Lessons 4–6 prove Theorem 1.1.

2. A biased block of primes

Suppose that (1.1) fails. By Mertens' estimate, $\sum_{T\le\log p\le2T}(\log p)/p=T+O(1)$. Each maximal bias is at most $1$, so there are a constant $\delta>0$ and an unbounded sequence of $T$ along which the primes with maximal bias at least $\delta$ have $(\log p)/p$-weight at least $c_\delta T$, with $c_\delta>0$. All limits below are taken along this sequence, and all constants may depend on $\delta$.

The scales. Put $\begin{gathered} \rho=\frac35,\\ \mu_0=10^{-6},\\ \gamma=10^{-7},\\ X=\exp\bigl(T^{1+\rho}\bigr), \end{gathered}$ and, as in Lesson 2, $A_X=A\cap[X^{9/10},X]$ and $D_X=D\cap[-X,-X^{9/10}]$. By Lemma 5.1 of Lesson 2, both have $\gg\sqrt X/(\log X)^3$ elements, and their uniform measures satisfy collision stability with $b=4$, for primes up to $\sqrt X/(\log X)^5$. Every prime used in this lesson and the next three satisfies $\log p\le2T$, far below this cutoff, because $\log\sqrt X=\frac12T^{1.6}$.

Pruning. The density term of Lemma 4.1 of Lesson 2 shows that the primes with $\sigma_p\notin[\frac13,\frac23]$ have $(\log p)/p$-weight $O(\log\log X)=O(\log T)$. For each prime $p$ with maximal bias at least $\delta$, choose a maximizing translate $t_p$ and a sign $\epsilon_p\in\{1,-1\}$ with $\begin{aligned} &\epsilon_p\,\mathbb E_{x\in S_p}\chi_p(x-t_p)\\ &\quad =\max_t\Bigl|\mathbb E_{x\in S_p}\chi_p(x-t)\Bigr|\ge\delta; \end{aligned}$ the mean is real because $\chi_p$ is real. Apply (5.2) of Lesson 2 to the functions $\phi_p(x)=\chi_p(x-t_p)$, for $A_X$ and for $D_X$. The primes at which either empirical mean differs from the corresponding uniform mean by more than $\delta/8$ have weight $O_\delta(\log T)$. Delete them and the unbalanced primes; weight $\gg T$ remains.

The opposite side. The complete sum $\sum_{x\in\mathbb F_p}\chi_p(x-t)$ vanishes, so $\begin{aligned} &\mathbb E_{x\in S_p^c}\chi_p(x-t)\\ &\quad =-\frac{\sigma_p}{1-\sigma_p}\,\mathbb E_{x\in S_p}\chi_p(x-t), \end{aligned}$ and $\sigma_p/(1-\sigma_p)\ge\frac12$ on balanced primes. At every remaining prime, therefore, $\begin{gathered} \epsilon_p\,\mathbb E_{x\in A_X}\chi_p(x-t_p)\ge\delta-\frac\delta8,\\ \epsilon_p\,\mathbb E_{x\in D_X}\chi_p(x-t_p)\le-\frac\delta2+\frac\delta8. \end{gathered}$

A dyadic block. The band $T\le\log p\le2T$ is covered by $O(T)$ intervals $[Z,2Z]$, and the remaining primes have weight $\gg T$, so some interval carries weight bounded below by a positive constant. Each prime in it has weight $(\log p)/p=O(\log Z/Z)$. We obtain:

Lemma 2.1. There are $Z$ with $\log Z\asymp T$ and a set $\mathcal P$ of primes in $[Z,2Z]$, with $J=|\mathcal P|\gg_\delta Z/\log Z$, such that $\frac13\le\sigma_p\le\frac23$ for $p\in\mathcal P$ and $\begin{gathered} \mathbb E_{x\in A_X}\epsilon_p\chi_p(x-t_p)\ge c_\delta,\\ \mathbb E_{x\in D_X}\epsilon_p\chi_p(x-t_p)\le-c_\delta\\ (p\in\mathcal P),\quad(2.1) \end{gathered}$ with $c_\delta=3\delta/8$. $\square$

Amplifier primes. Apply the density term of Lemma 4.1 of Lesson 2 to the band $T^\gamma\le\log p\le2T^\gamma$. Its total weight is $T^\gamma+O(1)$, and the unbalanced primes in it have weight $O(\log T)=o(T^\gamma)$. The same dyadic argument gives a block $[z,2z]$ with $\log z\asymp T^\gamma$ containing $\gg z/\log z$ primes $\ell$ with $\frac13\le\sigma_\ell\le\frac23$. Put $K=\Bigl\lfloor0.02\,\frac{\log Z}{\log(2z)}\Bigr\rfloor,$ and let $\mathcal L$ be the product of $K$ distinct such primes; there are enough of them, because $z/\log z\ge\exp(cT^\gamma)/T$ exceeds every fixed power of $T$. Let $k$ be the least even integer with $k\ge\log X/\log Z+10$. Then $\begin{gathered} K\asymp T^{1-\gamma},\\ \mathcal L\le Z^{0.02},\\ k\asymp T^\rho,\\ K=o(\log Z),\\ \frac Kk\gg T^{1-\gamma-\rho}.\quad(2.2) \end{gathered}$ Indeed $\log\mathcal L\le K\log(2z)\le0.02\log Z$, and $\log X/\log Z=T^{1.6}/\log Z\asymp T^{0.6}$. Since $2z<Z$, no prime factor of $\mathcal L$ lies in $\mathcal P$.

3. An amplified moment

For a prime $\ell\mid\mathcal L$ let $f_\ell=\mathbf 1_{S_\ell}-\sigma_\ell$, viewed as a function on $\mathbb Z$ of period $\ell$. Put $\begin{gathered} \lambda=\frac1{16},\\ W(n)=\prod_{\ell\mid\mathcal L}\bigl(1+\lambda f_\ell(n)\bigr),\\ H(n)=\frac1J\sum_{p\in\mathcal P}\epsilon_p\chi_p(n-t_p). \end{gathered}$ The weight $W$ is positive, has period $\mathcal L$, and has mean $1$ over a period: by the Chinese remainder theorem the residues of $n$ modulo the different $\ell$ are independent and uniform, and each $f_\ell$ has mean $0$. If $n\in A_X$, then $n>\ell+N_*$, so $n\bmod\ell\in S_\ell$ and $f_\ell(n)=1-\sigma_\ell\ge\frac13$. Hence $\begin{gathered} W(n)\ge(1+\lambda/3)^K\\ (n\in A_X).\quad(3.1) \end{gathered}$ The amplifier $W$ rewards integers that behave like elements of $A$ at the small primes $\ell$.

A smooth weight. Fix a real even smooth function $\varphi_0\ge0$ supported in $[-\frac18,\frac18]$, not identically zero, and let $g_0(x)=\int\varphi_0(y)e(xy)\,dy$. Then $g_0$ is real and even, and for $|x|\le1$, $g_0(x)\ge\cos(\pi/4)\int\varphi_0>0$. Put $\psi=g_0^2$. With the convention $\widehat\psi(y)=\int\psi(x)e(-xy)\,dx$, the transform $\widehat\psi=\varphi_0*\varphi_0$ is smooth, even, real and supported in $[-\frac14,\frac14]$, and $\psi$ is a nonnegative even Schwartz function bounded below by a positive constant on $[0,1]$.

Define $\begin{gathered} I=\sum_{n\in\mathbb Z}\psi(n/X)\,W(n)\,H(n)^k,\\ I_0=\sum_{n\in\mathbb Z}\psi(n/X)\,W(n). \end{gathered}$

Lemma 3.1. For large $T$, $\begin{gathered} I\ge\sqrt X\,e^{0.016K},\\ I_0=X\,\widehat\psi(0)\ll X.\quad(3.2) \end{gathered}$

Proof. All terms of $I$ are nonnegative, since $k$ is even. Restrict to $n\in A_X$, where $n/X\in[0,1]$ and $\psi(n/X)\gg1$. By (3.1) and Jensen's inequality for the convex function $x\mapsto x^k$, $\begin{aligned} I&\gg(1+\lambda/3)^K\sum_{n\in A_X}H(n)^k\\ &\ge(1+\lambda/3)^K\\ &\quad \,|A_X|\Bigl(\mathbb E_{n\in A_X}H(n)\Bigr)^k\\ &\ge(1+\lambda/3)^K\,\frac{\sqrt X}{(\log X)^3}\,c_\delta^k, \end{aligned}$ because $\mathbb E_{n\in A_X}H(n)\ge c_\delta$ by (2.1). Here $\log(1+\lambda/3)=\log(49/48)>0.0206$, while $k+\log\log X\ll T^{0.6}=o(K)$ by (2.2). So $\log I$$\ge\frac12\log X+0.0206K-o(K)$$\ge\frac12\log X+0.016K$ for large $T$.

For $I_0$, expand $W(n)=\sum_{h\bmod\mathcal L}\widehat W(h)\,e_{\mathcal L}(hn)$, where $\widehat W(0)=1$ is the mean. Poisson summation (Poisson summation, theta, and the functional equation, Theorem 1.2, applied to $x\mapsto\psi(x/X)e(hx/\mathcal L)$) gives $\begin{aligned} &\sum_n\psi(n/X)\,e_{\mathcal L}(hn)\\ &\quad =X\sum_{j\in\mathbb Z}\widehat\psi\bigl(X(j-h/\mathcal L)\bigr). \end{aligned}$ If $h\not\equiv0\pmod{\mathcal L}$, every argument has absolute value at least $X/\mathcal L>\frac14$, so the sum vanishes; for $h=0$ only $j=0$ survives. $\square$

4. Removing repeated primes

Expanding $H(n)^k$ produces products over $k$-tuples of primes from $\mathcal P$. The Poisson computation of Section 5 works for tuples of distinct primes. The following identity controls the rest.

Lemma 4.1 ([OpenAI-013, (3.6)]). Let $y_1,\dots,y_J\in\{0,1,-1\}$, put $H=J^{-1}\sum_iy_i$, and let $k\ge2$ be even. Then, for an absolute constant $C$, $\begin{aligned} &\Bigl|H^k-\frac1{J^k}\sum_{\substack{i_1,\dots,i_k\in[J]\\\text{distinct}}}y_{i_1}\cdots y_{i_k}\Bigr|\\ &\quad \le\sum_{1\le j\le k/2}\Bigl(\frac{k^C}J\Bigr)^j|H|^{k-2j}.\quad(4.1) \end{aligned}$

Proof. Write $p_m=\sum_iy_i^m$ for the power sums. The sum over distinct indices satisfies $\begin{aligned} &\sum_{\text{distinct}}y_{i_1}\cdots y_{i_k}\\ &\quad =\sum_{\sigma\in S_k}\operatorname{sgn}(\sigma)\prod_{c}p_{|c|},\quad(4.2) \end{aligned}$ where $c$ runs over the cycles of $\sigma$ and $|c|$ is the length. To see this, sum the right side over the permutations with a given set partition $\pi$ of $[k]$ into cycle supports. A block of size $m$ is the support of $(m-1)!$ cycles, each of sign $(-1)^{m-1}$, so the right side equals $\sum_\pi\prod_{B\in\pi}(-1)^{|B|-1}(|B|-1)!\;p_{|B|}$, and $\prod_{B\in\pi}p_{|B|}$ is the sum of $y_{i_1}\cdots y_{i_k}$ over the index tuples that are constant on each block of $\pi$. The coefficient of a tuple whose equality pattern is the partition $\tau$ is therefore $\sum_{\pi\le\tau}\prod_{B\in\pi}(-1)^{|B|-1}(|B|-1)!$, the sum over the partitions $\pi$ finer than $\tau$. This factorizes over the blocks of $\tau$: for a block of size $n$ it is $\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)$, which is $1$ for $n=1$ and $0$ for $n\ge2$. So exactly the tuples of distinct indices survive, each with coefficient $1$.

Because $y_i\in\{0,\pm1\}$, $p_m=JH$ for odd $m$, and $0\le p_m\le J$ for even $m$. The identity permutation contributes $(JH)^k$. Let $\sigma\ne\mathrm{id}$ have $k-2j$ cycles of odd length; the number of odd cycles has the parity of $k$, and $j\ge1$ because only the identity has $k$ odd cycles. Then $2j=\sum_{c\text{ odd}}(|c|-1)+\sum_{c\text{ even}}|c|,$ so the number of even cycles is at most $j$, and the term of $\sigma$ is at most $(J|H|)^{k-2j}J^{j}$ in absolute value. An odd cycle of length $m\ge3$ moves $m\le\frac32(m-1)$ points, and an even cycle moves $|c|$ points, so $\sigma$ moves at most $3j$ points. A permutation moving exactly $t$ given points can be chosen in at most $t!$ ways, so there are at most $\sum_{t\le3j}\binom kt t!\le(3j+1)k^{3j}\le k^{5j}$ permutations moving at most $3j$ points; the last step uses $3j+1\le4^j\le k^{2j}$. Summing over $j$ and dividing by $J^k$ gives (4.1) with $C=5$. $\square$

Application to the moment. Apply (4.1) at each $n$ with $y_p=\epsilon_p\chi_p(n-t_p)$, multiply by $\psi(n/X)W(n)\ge0$, and sum. By Hölder's inequality for the measure $\psi(n/X)W(n)$, $\begin{aligned} &\sum_n\psi(n/X)W(n)|H(n)|^{k-2j}\\ &\quad \le I^{1-2j/k}I_0^{2j/k}. \end{aligned}$ So the sum over distinct tuples differs from $I$ by at most $I\sum_{j\ge1}\bigl(k^CJ^{-1}(I_0/I)^{2/k}\bigr)^j$. By (3.2), $(I_0/I)^{2/k}\ll X^{1/k}e^{-0.032K/k}$$\le Ze^{-0.032K/k}$, because $k\ge\log X/\log Z$. Since $J\gg Z/\log Z$, $\begin{aligned} &\frac{k^C}J\Bigl(\frac{I_0}I\Bigr)^{2/k}\\ &\quad \ll k^C\log Z\;e^{-0.032K/k}\longrightarrow0, \end{aligned}$ because $K/k\gg T^{0.4-\gamma}$ while $\log(k^C\log Z)=O(\log T)$. Hence $\begin{aligned} &\frac1{J^k}\sum_n\psi(n/X)W(n)\\ &\qquad \sum_{\text{distinct}}\prod_{i=1}^k\epsilon_{p_i}\chi_{p_i}(n-t_{p_i})\\ &\quad =(1+o(1))\,I.\quad(4.3) \end{aligned}$

5. Poisson summation with arbitrary translations

Choose an ordered $k$-tuple of distinct primes of $\mathcal P$ uniformly at random. Every function used below depends on the tuple only through the product $M$ of its entries, and $\mathbb E_M$ denotes the expectation over the random tuple. Put $\begin{gathered} \chi_M=\prod_{p\mid M}\chi_p,\\ \epsilon_M=\prod_{p\mid M}\epsilon_p,\\ 0\le t_M<M,\\ t_M\equiv t_p\pmod p\ (p\mid M). \end{gathered}$ The character $\chi_M$ is the Jacobi symbol $(\cdot/M)$; it is real and primitive modulo the squarefree odd number $M$, as a product of primitive characters to coprime moduli. Define $\begin{gathered} R=\frac MX,\\ \theta=\frac{t_M}M,\\ h_0\equiv-t_M\overline M\pmod{\mathcal L},\\ 0\le h_0<\mathcal L, \end{gathered}$ where $\overline M$ is an inverse of $M$ modulo $\mathcal L$. Bars denote inverses modulo the modulus in the expression. Since $Z^k\le M\le(2Z)^k$ and $Z^{k-12}<X\le Z^{k-10}$ by the choice of $k$, $Z^{10}\le R\le Z^{12+o(1)}. \tag{5.1}$ The left side of (4.3) equals $(J)_k J^{-k}\,\mathbb E_M\,\epsilon_M\sum_n\psi(n/X)$$W(n)\chi_M(n-t_M)$, where $(J)_k=J(J-1)\cdots(J-k+1)$.

Local transforms of the amplifier. For $d\mid\mathcal L$ put $f_d=\prod_{\ell\mid d}f_\ell$ (a function modulo $d$ by the Chinese remainder theorem), $f_1=1$, and $g_d(b)=\sqrt d\;\mathbb E_{x\bmod d}\,f_d(x)\,e_d(-bx).$

Lemma 5.1. For $d\mid\mathcal L$, $\begin{gathered} \mathbb E_{b\bmod d}|g_d(b)|^2\le1,\\ |g_d(b)|\le\sqrt d,\\ g_d(b)=\prod_{\ell\mid d}g_\ell\bigl(b\,\overline{d/\ell}\bigr), \end{gathered}$ where $\overline{d/\ell}$ is an inverse of $d/\ell$ modulo $\ell$. If $d>1$, $g_d$ vanishes at every $b$ with $(b,d)>1$.

Proof. By Parseval, $\mathbb E_b|g_d(b)|^2=\sum_b|\widehat{f_d}(b)|^2$$=\mathbb E_x|f_d(x)|^2=\prod_\ell\sigma_\ell(1-\sigma_\ell)\le1$. The bound $|g_d|\le\sqrt d$ holds since $|f_d|\le1$. For coprime $\ell$ and $d/\ell$ one has $1/d\equiv\overline{d/\ell}/\ell+\overline\ell/(d/\ell)\pmod1$, and iterating over the prime factors of $d$ gives $bx/d\equiv\sum_{\ell\mid d}b\,\overline{d/\ell}\,x/\ell\pmod1$; the expectation over $x\bmod d$ then factors by the Chinese remainder theorem. If $\ell\mid(b,d)$, the factor at $\ell$ is $g_\ell(0)=\sqrt\ell\,\mathbb E f_\ell=0$. $\square$

The Poisson dual. Expand $W=\sum_{d\mid\mathcal L}\lambda^{\omega(d)}f_d$, where $\omega(d)$ is the number of prime factors. For fixed $d$, the function $F(n)=f_d(n)\chi_M(n-t_M)$ has period $Md$, and $(M,d)=1$. Poisson summation, as in the proof of Lemma 3.1, gives $\begin{gathered} \begin{aligned} &\sum_n\psi(n/X)F(n)\\ &\quad =X\sum_{u\in\mathbb Z}\widehat F(u)\,\widehat\psi\Bigl(\frac{u}{Rd}\Bigr), \end{aligned}\\ \widehat F(u)=\mathbb E_{n\bmod Md}F(n)\,e_{Md}(-un), \end{gathered}$ using that $\widehat\psi$ is even and $X/(Md)=1/(Rd)$. By the Chinese remainder theorem, with $un/(Md)$$\equiv u\overline Mn/d+u\overline dn/M\pmod1$, $\begin{aligned} \widehat F(u)&=\frac{g_d(u\overline M)}{\sqrt d}\\ &\quad \cdot\mathbb E_{y\bmod M}\chi_M(y-t_M)\\ &\quad \,e_M(-u\overline dy). \end{aligned}$ Substitute $y=z+t_M$. For the primitive character $\chi_M$, $\sum_z\chi_M(z)e_M(az)=\chi_M(a)\tau(\chi_M)$ for every integer $a$, where $\tau(\chi_M)$ is its Gauss sum, of modulus $\sqrt M$ (Gauss sums, Theorems 1.1 and 2.1; $\chi_M$ is real). Since $\chi_M(\overline d)=\chi_M(d)$, $\begin{aligned} \widehat F(u)&=\frac{\tau(\chi_M)}{M\sqrt d}\,\chi_M(-1)\chi_M(d)\\ &\quad \,\chi_M(u)\,g_d(u\overline M)\\ &\quad \,e_M(-u\overline dt_M). \end{aligned}$

The phase. The integer $t_M+Mh_0$ is divisible by $\mathcal L$, hence by $d$. Write $d\overline d+Mm'=1$ with an integer $m'\equiv\overline M\pmod d$. Then $\begin{aligned} \frac{\overline d\,t_M}M&=\frac{t_M}{dM}-\frac{m't_M}d\\ &\equiv\frac{\theta}d+\frac{h_0}d\pmod{\tfrac1d\mathbb Z}, \end{aligned}$ because $m't_M\equiv\overline Mt_M\equiv-h_0\pmod d$; multiplying by the integer $u$, $e_M(u\overline dt_M)=e\bigl((\theta+h_0)u/d\bigr)$.

Collect the terms, replace $u$ by $-u$ (using $\chi_M(-u)=\chi_M(-1)\chi_M(u)$ and that $\widehat\psi$ is even), divide by $\sqrt X$, and use $\sqrt X\,|\tau(\chi_M)|/(M\sqrt d)=1/\sqrt{Rd}$. The result is that $\epsilon_M X^{-1/2}\sum_n\psi(n/X)W(n)\chi_M(n-t_M)$ equals a complex number of modulus $1$, depending only on the tuple, times $\begin{aligned} &\sum_{d\mid\mathcal L}\frac{\lambda^{\omega(d)}\chi_M(d)}{\sqrt{Rd}} \sum_{u\in\mathbb Z}\chi_M(u)\\ &\quad \,g_d(-u\overline M)\,e\Bigl(\frac{(h_0+\theta)u}d\Bigr)\\ &\quad \,\widehat\psi\Bigl(\frac u{Rd}\Bigr).\quad(5.2) \end{aligned}$ The term $u=0$ vanishes since $\chi_M(0)=0$.

Positive frequencies. Let $\mathcal F_M$ be the part of (5.2) with $u>0$. Since $f_d$ is real, $\overline{g_d(b)}=g_d(-b)$; since $\widehat\psi$ is real and even, the part with $u<0$ equals $\chi_M(-1)\overline{\mathcal F_M}$. So (5.2) has modulus at most $2|\mathcal F_M|$. The normalization $(J)_k/J^k$ is at most $1$, and (4.3) and Lemma 3.1 give $(1+o(1))I/\sqrt X\ge e^{0.016K}/2$. Therefore $\mathbb E_M|\mathcal F_M|\ge e^{0.015K}. \tag{5.3}$

6. Squarefree kernels of the frequencies

Every integer $u\ge1$ has a unique factorization $\begin{gathered} u=svw^2,\\ s\ \text{squarefree},\ (s,\mathcal L)=1,\\ v\mid\mathcal L,\\ w\ge1: \end{gathered}$ $s$ is the product of the primes not dividing $\mathcal L$ that occur in $u$ to an odd power, $v$ the product of those dividing $\mathcal L$ that occur to an odd power, and $w^2=u/(sv)$. Then $\chi_M(u)=\chi_M(s)\chi_M(v)\mathbf 1_{(w,M)=1}$, and since $M$ is squarefree, $\mathbf 1_{(w,M)=1}=\sum_{P\mid(w,M)}(-1)^{\omega(P)}$.

For positive integers $P$, divisors $d,v$ of $\mathcal L$, and squarefree $s$ coprime to $\mathcal L$, define $\begin{aligned} N_{dsv}&=\sqrt{\frac{Rd}{sv}},\\ G_d^{(P)}(s,v)&=\frac1{N_{dsv}}\sum_{\substack{w\ge1\\P\mid w}}g_d(-svw^2\overline M)\,\\ &\qquad e\Bigl(\frac{(h_0+\theta)svw^2}d\Bigr)\,\\ &\qquad\widehat\psi\Bigl(\frac{svw^2}{Rd}\Bigr),\\ C_s^{(P)}&=\sum_{v\mid\mathcal L}\frac{\chi_M(v)}{\sqrt v}\sum_{d\mid\mathcal L}\lambda^{\omega(d)}\chi_M(d)\,\\ &\qquad G_d^{(P)}(s,v).\quad(6.1) \end{aligned}$ Since $1/\sqrt{Rd}=1/(N_{dsv}\sqrt{sv})$, regrouping (5.2) gives $\begin{aligned} \mathcal F_M&=\sum_{P\mid M}(-1)^{\omega(P)}\\ &\quad \sum_{\substack{s\ \text{squarefree}\\ (s,\mathcal L)=1}}\frac{\chi_M(s)}{\sqrt s}\,C_s^{(P)}.\quad(6.2) \end{aligned}$

Ranges. $\widehat\psi$ vanishes outside $[-\frac14,\frac14]$, so every nonzero term has $svw^2\le Rd/4\le R\mathcal L$. By (5.1) and (2.2), $R\mathcal L<Z^{14}$ for large $T$. Hence only $s<Z^{14}$ and $w<Z^7$ occur, and only $P\le Z^7$. A divisor of $M$ of size at most $Z^7$ has at most $7$ prime factors, since each prime factor is at least $Z$; so (6.2) has $O(k^7)$ values of $P$, and the inner sums are finite.

The plan. For many tuples, (5.3) forces a large value of one of the sums $G_d^{(1)}(s,v)$ with $s$ small. On an arithmetic progression $w=x+dj$, the phase of $G_d^{(1)}(s,v)$ is a quadratic polynomial in $j$ with leading coefficient $svd\theta$. A large quadratic exponential sum forces this coefficient to be close to a rational number with small denominator, which relates $\theta=t_M/M$, and hence the translates $t_p$, across the primes of $M$. Since the arrays $C_s^{(P)}$ depend on the random tuple, the moment estimate of Lesson 4 is proved uniformly over a fixed family of arrays that contains their discretizations.

7. Exercises

7.1. Verify (4.2) directly for $k=2$ and $k=3$: $\sum_{i\ne j}y_iy_j=p_1^2-p_2$ and $\sum_{\text{distinct}}y_iy_jy_l=p_1^3-3p_1p_2+2p_3$.

7.2. Let $M$ and $d$ be coprime positive integers, $\overline M$ an inverse of $M$ modulo $d$ and $\overline d$ an inverse of $d$ modulo $M$. Show that $\frac1{Md}\equiv\frac{\overline M}d+\frac{\overline d}M\pmod1$.

7.3. Show that for odd squarefree $M$, the character $\chi_M=\prod_{p\mid M}\chi_p$ is primitive modulo $M$, and that $\chi_M(-1)=(-1)^{(M-1)/2}$.

7.4. In the construction of $\psi$, show that $\widehat\psi=\varphi_0*\varphi_0$, and that $\psi$ decays faster than any power of $|x|$.

8. Solutions

7.1. For $k=2$ the permutations are the identity, with term $p_1^2$, and the transposition, with sign $-1$ and one cycle of length $2$, giving $-p_2$. The left side is $\bigl(\sum y_i\bigr)^2-\sum y_i^2$. For $k=3$: the identity gives $p_1^3$; the three transpositions give $-p_1p_2$ each; the two $3$-cycles are even and give $p_3$ each. Directly, $p_1^3=\sum_{\text{distinct}}+3\sum_{i\ne j}y_i^2y_j+p_3$, and $\sum_{i\ne j}y_i^2y_j=p_2p_1-p_3$, so $\sum_{\text{distinct}}=p_1^3-3p_1p_2+2p_3$.

7.2. The number $1-d\overline d$ is divisible by $M$ and $1-M\overline M$ by $d$, so $N=1-d\overline d-M\overline M$ is divisible by both, hence by $Md$. Then $\frac{\overline M}d+\frac{\overline d}M=\frac{M\overline M+d\overline d}{Md}=\frac{1-N}{Md}$$\equiv\frac1{Md}\pmod1$.

7.3. Each $\chi_p$ is a nonprincipal character modulo the prime $p$, hence primitive. If $\chi_M$ were induced from a proper divisor $M'$ of $M$, choose $p\mid M$ with $p\nmid M'$ and $a\equiv1\pmod{M/p}$ with $\chi_p(a)=-1$; then $a\equiv1\pmod{M'}$ but $\chi_M(a)=-1$. For the value at $-1$: $\chi_p(-1)=(-1)^{(p-1)/2}$, and $\sum_{p\mid M}\frac{p-1}2\equiv\frac{M-1}2\pmod2$ because $\frac{ab-1}2-\frac{a-1}2-\frac{b-1}2=\frac{(a-1)(b-1)}2$ is even for odd $a,b$.

7.4. The transform of $g_0$ is $\varphi_0$ by Fourier inversion, and the transform of a product is the convolution of the transforms. Since $\widehat\psi$ is smooth with compact support, repeated integration by parts in $\psi(x)=\int\widehat\psi(y)e(xy)\,dy$ gives $|\psi(x)|\ll_j(1+|x|)^{-j}$ for every $j$, and the same for all derivatives.

Sources

[OpenAI-013] OpenAI, The additive indecomposability of the primes, OpenAI Math Release preprint, 24 September 2026, Section 3: Proposition 3.1, the biased block, the amplified moment, (3.6)–(3.12), and Corollary 3.2. https://github.com/openai/math/tree/main/preprints/the-additive-indecomposability-of-the-primes-September-24-2026