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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
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The congruence $n\equiv\varphi(n) \pmod p$
Has the congruence $n\equiv\varphi(n) \pmod p$, with $p$ being an odd prime not dividing $n$, been examined before?
Because it is easy to find solutions for $n$ with few primes in its decomposition ( …
1
vote
0
answers
197
views
Impossibility of certain carmichael numbers
T. Wright http://www.math.jhu.edu/~wright/carmichael.pdf proves that $lcm ((p_1-1),(p_2-1),...)$ over the prime factors of a carmichael number cannot be of the form $2^s$
How to prove that carmichael …
1
vote
1
answer
333
views
Generalized Lehmer Euler Conjecture
Has the following generalized version of the Lehmer's conjecture for the Euler totient function
(original version: there are no composite solutions to the equation $n-1 \equiv 0 (\varphi(n))$)
Find c …
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0
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35
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Is there any characterization of cototients equal to prime powers?
Do we know something about numbers $x$ with
$x-\varphi(x)=p^n$ for some $n$ and prime $p$
Are there any for each prime $p$?
2
votes
1
answer
657
views
Inequality with Euler's totient function
Numerical analysis of the first several hundred n suggests the following inequality:
$\varphi(3^n-2) \ge 2\cdot3^{n-1}$
2
votes
1
answer
126
views
another generalized Lehmer-Euler-conjecture
Lehmer-Euler conjecture states, that there a no composite numbers $n$ satisfying
$(n-1) \equiv 0(\varphi(n))$
Are there any results out there for
$a(n-1) \equiv 0(\varphi(n))$, for integer $a$
esp.
…
4
votes
1
answer
249
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Proof in Schemmel's Paper
I tried to reproduce the results in a paper of Schemmel (where Schemmel's totients were introduced)
$n$ consecutive numbers is an arithmetic sequence of numbers with common difference $n$ (e.g. 1,5, …
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1
answer
102
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Are there any solutions to this congruence system
Let $p,q_i,i=1,2,..m$ be odd primes with integer $m>2$
Does this system of congruences have any solutions?
$\prod_{i=1}^m(q_i-1)\equiv2(p^2)$
$\prod_{i=1}^mq_i\equiv2(p^3)$
6
votes
2
answers
494
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Are twin primes the only solution to this equation?
Let $p,q_i, i \ge 1$ be primes, $m$ a positive integer.
The equation
$$
p.\prod_{i=1}^m(q_i-1)-(p-1).\prod_{i=1}^mq_i=2
$$
for $m=1$ has all twin primes $p,q_1=p+2$ as solution.
Are there solutions …