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I wonder whether you mean something else, but if you want to know what the answer is (dimensions of spaces, Hecke action etc) then all of this can be read off from Jacquet-Langlands and the analogous …

No it's not true. An equivalent question is whether the product of $(1-1/p)$, where $p$ ranges over all the primes for which 2 has odd order mod $p$, is greater than or equal to $1/2$. However an expl …

As Gerhard says, $m=1$ works giving $p=5$ and $k=3$. I claim that this is the only time it happens. Let $m\geq2$ be an integer, and set $p=2^{2m+1}-2^{m+1}+1$. I don't care if $p$ is prime or not, bu …

The case $n=4$ and $m=35$ given in both the question and the original German paper (link now removed from question) is a little misleading. In this case the sets are disjoint. The case $n=2$ and $m=35 …

No. How about $a=9$ and $b=16$? Then $c=25$ so $\Omega(a)=\Omega(c)=2\leq\Omega(b)$, the gcd's are $3,16,5$ so the left hand side is 6 and the right hand side only 5. Edit: if $a=316$ and $b=27$ then …

Greg notes that $(Z/pZ)[[x]]$ is a limit of the integers of a sequence of totally ramified extensions of $Q_p$. At the back of my mind I wonder whether he'd be interested by the Fontaine-Wintenberger …

Firstly, at the beginning of the question you are missing irreducibility/cuspidality assumptions. If $\rho_1$ and $\rho_2$ are $\ell$-adic Galois reps with the same char poly in a set of primes of den …

Here's my guess: it might be out of reach to prove that $g_n$ tends to infinity, but it probably does, because $1!+2!+\ldots+(p-1)!$ is a "random" number mod $p$, so the chances that it's divisible by …

I don't think I even know a natural non-zero map from the space $S_2(\Gamma_1(N);\mathbf{Q}_p)$ ("classical" modular forms with $p$-adic coefficients, defined for example as global sections of an appr …

This answer pertains to Michael's comment above. Let $c$ be $e^{\pi\sqrt{163}}$. It is well-known that $c$ is close to an integer. More precisely, $c$ is about $10^{-12}$ from an integer of size abou …

I absolutely agree that Deligne is very terse. One thing that I ultimately found very helpful is Carayol's two papers where he proves the analogous theorem for Hilbert modular forms. I say "ultimately …

If I've understood the question correctly, this is just a Jacobi sum. The question only seems to make sense for $p$ a prime congruent to 1 mod 3. In this case, the order 3 character and the quadratic …

Not an answer, but too long for a comment. I can't see any tricks to do this other than a brute force computational approach. The naive approach would be to loop from $2^{96}$ to $2^{128}$ squaring e …

$\newcommand{\Q}{\mathbf{Q}}\newcommand{\Z}{\mathbf{Z}}$ I now suspect the answer might be no! This isn't a complete answer but it might be an idea that turns into one. So let me assume that such $p$ …

Greg, I want to say some basic things, but people are giving quite "high-brow" answers and what I want to say is a bit too big to fit into a comment. So let me leave an "answer" which is not really an …