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You can just compute the spectrum exactly using the formula mentioned by Benjamin Steinberg: $$\sum_g \operatorname{ord}(g) \frac{\chi(g)}{\chi(1)}.$$ There are two or four characters of degree $1$ ac …

Note that $\mathcal{A}$ is contained in $\def\F{\mathbb{F}}\F_{q^2}$ if and only if $(q^n+1)/(q+1)$ divides $q^2-1$, and it is easy to check with a bit of case analysis that this happens only if $(n, …

Extended comment, generalizing @IlyaBogdanov's comment about $2n-1$. Fix $n$ and let $$x_m = a(m, n) + n - m - 1.$$ Then $(x_m)$ obeys the similar recurrence $$ x_m = x_{m-1} + \gcd(x_{m-1}, n-m) - …

Yes, the estimate is uniform in $a$ and $d$, I think. It suffices to handle the two-dimensional case, by considering two-dimensional slices. So consider the circle $(x - a)^2 + (y-b)^2 = R^2$ with $a, …

It is basic that the norm map $N:\mathbf{F}_{q^n}^* \to \mathbf{F}_q^*$ is surjective for finite fields. In fact $N(x) = x^{(q^n-1)/(q-1)}$. How well does this simple fact extend to subspaces? A basi …

Naturally, this was also considered by Erdös and Graham. Graham mentions at the top of page 10 here for instance: http://www.math.ucsd.edu/~ronspubs/13_03_Egyptian.pdf. I'm not aware of any progress.

The difference cover problem has been better studied in the context of $\mathbf{Z}$. Redei, Renyi, and others in the 40s asked for the size of the smallest set $A$ such that $A-A$ covers $\{1,2,\dots, …

$\def\Z{\mathbf{Z}}$If $N$ is prime and $f(x)$ has the form $\zeta^{g(x)}$, where $\zeta$ is an $N$th root of unity and $g$ is some function from $\Z/N\Z$ to $\Z$, then $g$ must be quadratic. The foll …

Your argument for three exponentials can be simplified a bit by using the multiplicative version of van der Corput instead of the additive version. Specifically, if your equation $$2^{y^{2^k}} = 2^{y} …

We can reduce Lucia's upper bound of $3/8$ a little further as follows. Begin by taking the $n/4$ geometric progressions of common ratio $2$ beginning at each odd number at most $n/2$. Then for each o …

Take $s_n = n^2$ and $\alpha$ irrational.