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The version you stated is certainly not correct. A smooth functions supported on $[1,2]$ vanishes at $1$ and $2$, so the sum you write vanishes whenever $m$ is a power of $2$. Rather the trick require …

If you're OK with "almost every", we can use a $L^2$ method and multiplicative character sum bounds to do better. Let's estimate $$ \sum_{c \in \mathbb F_p^\times} \left( \left[\sum_{a,b \in I | ab=c …

I think one can get a lower bound more like $x_n \geq e^{ c \sqrt{n \log n}}$, or equivalent $p_{n+1} = O ( (\log x_n)^2)$, conditional on GRH. One simply reverses the sign in the argument giving the …

For general tori this problem gets really weird. For instance, given any polynomial equation $f(x_1,..,x_n)$ defined over $\mathbb F_p$, I can encode the point-counting problem $|\{x_1,\dots,x_n \in …

Let me try to explain why this problem is hard (concurring with Alexey Ustinov here). Fourier analysis shows that $$ N_p = \frac{1}{p^2} \sum_{m=0}^{p-1} \sum_{n=0}^{p-1} \left( \sum_{x=1}^{p-1} e_p( …

The proportion is 1. In fact, there is a nonempty Zariski open set such that all polynomials in the open set satisfy the condition. Indeed, a sufficient condition, for a polynomial of degree $n$ is th …

It is very likely that there are only finitely many primes with this property. Heuristically, the probability of this happening for the $N$th prime is $2^{2-N}$, and $\sum_{N \geq 2} 2^{2-N} =2$, so …

Let $d = \deg f$. Consider the covering of $\mathbb A^1_\mathbb Q$, with coordinate $s$, defined by $f(x_1) =s+1 ,\dots, f(x_t) =s +t$. The monodromy representation of this covering gives a map from t …

One approach is to use the Polya-Vinogradov method: i.e. using a Fourier transform mod $2q$, find $c_m$ such that $$\sum_{m=1}^{2q} c_m e \left( \frac{mn}{2q} \right)= \begin{cases} 1 & \textrm{if } 1 …

By Poisson: $$ \sum_{n \in \mathbb Z} f(n) e(\alpha n)= \sum_{m \in \mathbb Z} \hat{f} ( 2 \pi m + \alpha) $$ By the formula for the derivative of the Fourier transform: $$= \sum_{m \in \mathbb Z} \ …

Suppose $n=2m$ and $a=m+1$. Clearly $(a,m)=1$. Then there are no $x,y$ odd with $ax \equiv y$ mod $n$, $x<m$, $y<m$. Indeed in this case $$y \equiv ax= mx+x \equiv m+x \mod 2m$$ Clearly this cannot …

The different bounds you have given carry with them different assumptions. The first bound is stated for $\gcd(m,n,c)=1$. The second bound is stated when $a$ and $b$ are both prime to $p$ (though it …

There is a $48$-element symmetry group of the space of solutions. This by itself will create the appearance of the patterns when the number of solutions divided by $48$ is small. Imagine choosing $k$ …

$$\int_x^\infty \{ u\} u^{-s-1} du = \int_x^\infty \left( \{ u\}-\frac{1}{2}\right) u^{-s-1} du + \frac{1}{2} \frac{x^{-s}}{s}$$ and by integration by parts, $$\int_x^\infty \left( \{ u\}-\frac{1}{2} …

It's not completely clear from your question if you want lower or upper bounds. For lower bounds, the Odlyzko bounds come from analysis of the $L$-function. The lower bound we get by analyzing the $L …