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An element $x$ of $\bar{\mathbb{F}_p}$ satisfies $$\sum_{k=0}^{M}{M \choose k}^2x^k=0$$ iff the elliptic curve given by the equation $Y^2 = X(X-1)(X-x)$ is supersingular : see the paragraph on the Leg …

Yes. If we set $\alpha = (a+b)/2$ and $\beta=(a-b)/2$, then the lattice generated by the four vectors $$ \binom{\alpha}{\beta},\binom{-\beta}{\alpha},\binom{\alpha+\beta}{\alpha -\beta},\binom{\beta …

Let $p$ be a prime with $p \equiv 3 \mod 4$, $p \neq 3$, with its associate Legendre symbol $\left( \frac{.}{p} \right)$, and let $h$ be the class number of $\mathbb{Q} ( i \sqrt{p} )$. If $x \mapsto …

This is also directly related to the size of $\sum_{n \leq X} \mu(n)$ as follows : Assume that the estimate $\sum_{n \leq X} \frac{\mu(n)}{n} = O(\varepsilon(X))$ holds for some non increasing functi …

Most of the proofs of Dirichlet's theorem on primes in arithmetic progressions actually give a Mertens-like theorem, and then the (weaker) statement Chebyshev-like bound : if $(a,q) = 1$ then …

Let $x_j = \sum_{n \geq 1} \mathrm{1}_{a_n=j} 3^{-n}$. Then $$ x_0 + x_1 + x_2 = \frac{1}{2} \\ x_1 + 2x_2 = x \\ x_0 + 2x_1 = f(x). $$ If $x$ and $f(x)$ are algebraic then the equations above would …

I can show the (exact) inequality $$(*) \ \ \ \ \ \ V(Z_1) \geq \left(\frac{Z_1}{Z_2} \right)^2 V(Z_2) \ \ \quad (\frac{2}{A} \leq Z_1 \leq Z_2 \leq \frac{A}{2}), $$ for a smoothed version of $U$ def …

Let $K_n$ be the sets of vectors $x \in \mathbb{Z}^d $ with each coordinates $x_i$ between $1$ and $n$. For any subset $A$ of $K_n$, let $S(A)$ be the set of points $x \in K_n$ which are on some line …

By juan's answer above, one only needs to show $\sum_{p \leq e^{\frac{2}{\delta}}} \frac{1}{p^{1-\delta}} \leq \log \frac{1}{\delta} + O(1) $. But by estimating $p^{\delta} = 1 + O( \delta \log p )$, …

Your guess is correct. Since $2^{\omega(n)} = \sum_{d | n} \mu^2(d)$ one has $$ S = \sum_{(a,b,c) \in R(X)} 2^{\omega(4ac-b^2)} = \sum_{d \leq X} \mu^2(d) |R_d(X)| $$ where $R_d(X)$ is the set of inte …

Of course since $R = \{ (a,b) \in \mathbb{N} \times \mathbb{N} \ | \ (a=1) \implies (b=1) \} $. Indeed for $a > 1$ one can take $n = a^k-b$ for $k$ large enough.

One has $S_a(x) \sim C(a) x (\log x)^{2^a -1}$ where $$ C(a) = \Gamma(2^a)^{-1} \prod_p \left( 1 - \frac{1}{p} \right)^{2^a} \left( \sum_{k \geq 0} \frac{(k+1)^a}{p^k}\right). $$ This follows for exam …

One can take $r = 1 + \frac{d(d-1)}{2}$. Indeed, one can consider the map $$ (\alpha_i)_{i=1}^r \in [|1,N|]^{r} \mapsto (\sum_{i=1}^r \alpha_i^k)_{k=1}^{d-1} \in \prod_{i=1}^{d-1} [|1,rN^k|]. $$ The …

Using the decomposition $$ H(x) := \sum_{a \leq x} \frac{1}{a} = \log(x) + \gamma - \frac{\psi(x)}{x} + \int_{x}^{+\infty} \frac{\psi(t) d t}{t^2}, $$ where $\psi(t) = \{ t \} - \frac{1}{2}$, one ge …

You can simplify Ramachandra's method by bounding the last sum of p.305 using Brun-Titchmarsh inequality (or Montgomery & Vaughan error-term free version of it) instead of Van der Corput's method + Se …