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The number $S_q$ depends on the choice of additive character $\psi$: changing $\psi$ multiplies each $g(\psi,\chi)$ by some $(q-1)$-th root of unity depending also on $\chi$; this is often harmless, b …

No, it is not true that $M(D) = o(|D|^{3/2})$. For example, if $D = -4abcd$ where $a,b,c,d$ are odd, pairwise coprime and of roughly equal size, then the forms $ab\,x^2+cd\,y^2, ac\,x^2+bd\,y^2, ad\, …

The $n=4$ sum is accessible as follows. Expand $(j+k)^4$; by symmetry the odd terms $4j^3k$ and $4jk^3$ cancel out so we need only sum $(j^4 + 6j^2k^2 + k^4) / (j^2+k^2)^4$. You say you already know …

If $\zeta_K(1/2) \neq 0$ then $\zeta'_K(1/2) = 0$ if and only if $$ \log |D_K| = (\log(8\pi) + \gamma) n + \frac\pi2 r_1, $$ where $D_K$ is the discriminant of $K$, and $\gamma = 0.5772156649\ldots$ …

If $\Phi(0) = 1 + \epsilon$ then $\,\widehat{\!\Phi\!}\,(t) \gg \epsilon^{-1/2}$, even under the weaker assumption that $\,\widehat{\!\Phi\!}\,(t) \geq 0$ for all $t$; and this is best possible up to …

For starters we have $\,\widehat{\!\Phi\!}\,(0) \geq 2$. Indeed by Poisson summation $$ \sum_{m = -\infty}^\infty \Phi(m+1/2) = \sum_{n = -\infty}^\infty (-1)^n \,\widehat{\!\Phi\!}\,(n). $$ The LHS i …

We must assume that $1-\rho$ is not exceptional either. Then both $|\rho|$ and $|1-\rho|$ are $\gg 1 / \log p$, and the desired inequality follows, for instance because $$ \frac1{\rho (1-\rho)} = \fr …

Such attempted generalizations of ABC to four or more variables often fail to specializations of the identity $$ (x^2+xy-y^2)^3 + (x^2-xy-y^2)^3 = 2 (x^6 - y^6). \label{1}\tag{*} $$ One can use ellipt …

For any real $A \neq 1$ we have $$ \sum_{n \leq x} (\omega(n) - A \log \log x)^2 \sim (1-A)^2 x (\log \log x)^2 $$ as $x \to \infty$. This is a consequence of the quoted result $$ \sum_{n \leq x} (\o …

Yes, the minimal $\|(u,v,z)\|_\infty$ is within a constant factor of $\sqrt{|c|}$ (equivalently, of $\sqrt{\max(|a|,|b|)}$. The orthogonal complement of $(a,b,c) = (m^2-n^2, 2mn, m^2+n^2)$ contains …

You're surely right that there cannot be a "closed form" for such a series; but it can still be approximated to any desired precision. The defining sum $$ x = x(a) = \sum_{i=0}^\infty \sum_{j=0}^\in …

Here's an elementary proof of the inequality $$ (1) \qquad\qquad \sum_{k=1}^{N-1} \left(1-\frac{k}{N}\right)B_2(\{kx\}) \ge \frac1{12N} - \frac1{12}. \qquad\qquad\phantom{(1)} $$ This is nearly the …

Once you know that the coefficients are all positive (see postscript), it's easy to get an effective upper bound that grows as $\exp(4\pi \sqrt{n})$, which is within a factor $O(\sqrt n)$ of the corre …

Yes, it is true. This generating function $\sum_n a_n q^n$ turns out to be the same as $(3F(q^3)-F(q))/2$: they coincide through the $q^{100}$ term, which is more than enough to prove equality betwee …

You are right to question this. The product $\prod_p \left(1 - \chi(p)/p\right)^{-1}$ (where $\chi = (-1/\cdot)$ is the Dirichlet character mod $4$) does converge, and the limit is $L(1,\chi) = \pi/4 …