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GH from MO
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I think the sought result follows from Vinogradov's theorem. By Weyl's criterion and the orthogonality of Dirichlet characters, the sought result can be reformulated as follows. For every nonzero integer $k$, and for every Dirichlet character $\chi$ modulo $q$, we have $$\sum_{p<x}\chi(p)e(k\alpha p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ The function $n\mapsto\chi(n)$ can be written as a linear combination of additive characters $n\mapsto e((a/q)n)$ with $a\in\mathbb{Z}$, hence it suffices to show that $$\sum_{p<x}e((a/q+k\alpha)p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ This in turn follows from the equidistribution of $\{(a/q+k\alpha)p\}$, because $a/q+k\alpha$ is an irrational number. QED

The error terms will be similar as in Vinogradov's theorem, but with additional constants depending on $q$. The rate of convergence will depend on how well-approximable $\alpha$ iscan be approximated by rational numbers. See Chapter XI and the subsequent notes in Vinogradov's "The method of trigonometrical sums in the theory of numbers". I am sure there is a modern reference, but I am no expert in this subject.

I think the sought result follows from Vinogradov's theorem. By Weyl's criterion and the orthogonality of Dirichlet characters, the sought result can be reformulated as follows. For every nonzero integer $k$, and for every Dirichlet character $\chi$ modulo $q$, we have $$\sum_{p<x}\chi(p)e(k\alpha p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ The function $n\mapsto\chi(n)$ can be written as a linear combination of additive characters $n\mapsto e((a/q)n)$ with $a\in\mathbb{Z}$, hence it suffices to show that $$\sum_{p<x}e((a/q+k\alpha)p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ This in turn follows from the equidistribution of $\{(a/q+k\alpha)p\}$, because $a/q+k\alpha$ is an irrational number. QED

The error terms will be similar as in Vinogradov's theorem, but with additional constants depending on $q$. The rate of convergence will depend on how well-approximable $\alpha$ is by rational numbers. See Chapter XI and the subsequent notes in Vinogradov's "The method of trigonometrical sums in the theory of numbers". I am sure there is a modern reference, but I am no expert in this subject.

I think the sought result follows from Vinogradov's theorem. By Weyl's criterion and the orthogonality of Dirichlet characters, the sought result can be reformulated as follows. For every nonzero integer $k$, and for every Dirichlet character $\chi$ modulo $q$, we have $$\sum_{p<x}\chi(p)e(k\alpha p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ The function $n\mapsto\chi(n)$ can be written as a linear combination of additive characters $n\mapsto e((a/q)n)$ with $a\in\mathbb{Z}$, hence it suffices to show that $$\sum_{p<x}e((a/q+k\alpha)p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ This in turn follows from the equidistribution of $\{(a/q+k\alpha)p\}$, because $a/q+k\alpha$ is an irrational number. QED

The error terms will be similar as in Vinogradov's theorem, but with additional constants depending on $q$. The rate of convergence will depend on how well $\alpha$ can be approximated by rational numbers. See Chapter XI and the subsequent notes in Vinogradov's "The method of trigonometrical sums in the theory of numbers". I am sure there is a modern reference, but I am no expert in this subject.

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GH from MO
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I think the sought result follows from Vinogradov's theorem. By Weyl's criterion and the orthogonality of Dirichlet characters, the sought result can be reformulated as follows. For every nonzero integer $k$, and for every Dirichlet character $\chi$ modulo $q$, we have $$\sum_{p<x}\chi(p)e(k\alpha p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ The Dirichlet characterfunction $\chi$$n\mapsto\chi(n)$ can be written as a linear combination of additive characters $t\mapsto e((a/q)t)$$n\mapsto e((a/q)n)$ with $a\in\mathbb{Z}$, hence it suffices to show that $$\sum_{p<x}e((a/q+k\alpha)p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ This in turn follows from the equidistribution of $\{(a/q+k\alpha)p\}$, because $a/q+k\alpha$ is an irrational number. QED

The error terms will be similar as in Vinogradov's theorem, but with additional constants depending on $q$. The rate of convergence will depend on how well-approximable $\alpha$ is by rational numbers. See Chapter XI and the subsequent notes in Vinogradov's "The method of trigonometrical sums in the theory of numbers". I am sure there is a modern reference, but I am no expert in this subject.

I think the sought result follows from Vinogradov's theorem. By Weyl's criterion and the orthogonality of Dirichlet characters, the sought result can be reformulated as follows. For every nonzero integer $k$, and for every Dirichlet character $\chi$ modulo $q$, we have $$\sum_{p<x}\chi(p)e(k\alpha p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ The Dirichlet character $\chi$ can be written as a linear combination of additive characters $t\mapsto e((a/q)t)$ with $a\in\mathbb{Z}$, hence it suffices to show that $$\sum_{p<x}e((a/q+k\alpha)p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ This in turn follows from the equidistribution of $\{(a/q+k\alpha)p\}$, because $a/q+k\alpha$ is irrational. QED

The error terms will be similar as in Vinogradov's theorem, but with additional constants depending on $q$. The rate of convergence will depend on how well-approximable $\alpha$ is by rational numbers. See Chapter XI and the subsequent notes in Vinogradov's "The method of trigonometrical sums in the theory of numbers". I am sure there is a modern reference, but I am no expert in this subject.

I think the sought result follows from Vinogradov's theorem. By Weyl's criterion and the orthogonality of Dirichlet characters, the sought result can be reformulated as follows. For every nonzero integer $k$, and for every Dirichlet character $\chi$ modulo $q$, we have $$\sum_{p<x}\chi(p)e(k\alpha p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ The function $n\mapsto\chi(n)$ can be written as a linear combination of additive characters $n\mapsto e((a/q)n)$ with $a\in\mathbb{Z}$, hence it suffices to show that $$\sum_{p<x}e((a/q+k\alpha)p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ This in turn follows from the equidistribution of $\{(a/q+k\alpha)p\}$, because $a/q+k\alpha$ is an irrational number. QED

The error terms will be similar as in Vinogradov's theorem, but with additional constants depending on $q$. The rate of convergence will depend on how well-approximable $\alpha$ is by rational numbers. See Chapter XI and the subsequent notes in Vinogradov's "The method of trigonometrical sums in the theory of numbers". I am sure there is a modern reference, but I am no expert in this subject.

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GH from MO
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I think the sought result follows from Vinogradov's theorem. By Weyl's criterion and the orthogonality of Dirichlet characters, the sought result can be reformulated as follows. For every nonzero integer $k\in\mathbb{Z}$$k$, and for every Dirichlet character $\chi$ modulo $q$, we have $$\sum_{p<x}\chi(p)e(k\alpha p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ The Dirichlet character $\chi$ can be written as a linear combination of additive characters $t\mapsto e(ta/q)$$t\mapsto e((a/q)t)$ with $a\in\mathbb{Z}$, hence it suffices to show that $$\sum_{p<x}e((a/q+k\alpha)p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ This in turn follows from the equidistribution of $\{(a/q+k\alpha)p\}$, because $a/q+k\alpha$ is irrational. QED

The error terms will be similar as in Vinogradov's theorem, but with additional constants depending on $q$. The rate of convergence will depend on how well-approximable $\alpha$ is by rational numbers. See Chapter XI and the subsequent notes in Vinogradov's "The method of trigonometrical sums in the theory of numbers". I am sure there is a modern reference, but I am no expert in this subject.

I think the sought result follows from Vinogradov's theorem. By Weyl's criterion and the orthogonality of Dirichlet characters, the sought result can be reformulated as follows. For every $k\in\mathbb{Z}$, and for every Dirichlet character $\chi$ modulo $q$, we have $$\sum_{p<x}\chi(p)e(k\alpha p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ The Dirichlet character $\chi$ can be written as a linear combination of additive characters $t\mapsto e(ta/q)$, hence it suffices to show that $$\sum_{p<x}e((a/q+k\alpha)p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ This in turn follows from the equidistribution of $\{(a/q+k\alpha)p\}$, because $a/q+k\alpha$ is irrational. QED

The error terms will be similar as in Vinogradov's theorem, but with additional constants depending on $q$. The rate of convergence will depend on how well-approximable $\alpha$ is by rational numbers. See Chapter XI and the subsequent notes in Vinogradov's "The method of trigonometrical sums in the theory of numbers". I am sure there is a modern reference, but I am no expert in this subject.

I think the sought result follows from Vinogradov's theorem. By Weyl's criterion and the orthogonality of Dirichlet characters, the sought result can be reformulated as follows. For every nonzero integer $k$, and for every Dirichlet character $\chi$ modulo $q$, we have $$\sum_{p<x}\chi(p)e(k\alpha p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ The Dirichlet character $\chi$ can be written as a linear combination of additive characters $t\mapsto e((a/q)t)$ with $a\in\mathbb{Z}$, hence it suffices to show that $$\sum_{p<x}e((a/q+k\alpha)p)=o(\pi(x))\qquad\text{as}\qquad x\to\infty.$$ This in turn follows from the equidistribution of $\{(a/q+k\alpha)p\}$, because $a/q+k\alpha$ is irrational. QED

The error terms will be similar as in Vinogradov's theorem, but with additional constants depending on $q$. The rate of convergence will depend on how well-approximable $\alpha$ is by rational numbers. See Chapter XI and the subsequent notes in Vinogradov's "The method of trigonometrical sums in the theory of numbers". I am sure there is a modern reference, but I am no expert in this subject.

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GH from MO
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