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Results tagged with nt.number-theory
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user 2384
Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
3
votes
Pell Numbers and the Primes
Your claim is true. Note that we can solve the recurrences to give
$$a_p=\frac{1}{\sqrt{2}}\left((1+\sqrt{2})^{p+1}-(1-\sqrt{2})^{p+1}\right)$$
for the first sequence. So for a prime $p$, we have that …
- 84.6k
5
votes
Why are the only numbers $m$ for which $n^{m+1}\equiv n \bmod m$ also the only numbers such ...
I don't see how to find all $m$ for which $\sum_{k=1}^m k^m \equiv 1 \bmod{m}$, which seems to be difficult, but you ask only why do 1, 2, 6, 42, and 1806 work. One answer is that if you plug them in …
- 84.6k
6
votes
Inequality with Euler's totient
As already mentioned in the comments, one cannot expect inequalities of the form $$\varphi(an+b)>\varphi(cn+d)$$
(provided $ad\neq bc$) to hold for all n. In fact, as is proved here for any polynomial …
- 84.6k
5
votes
Accepted
In the sense of Collatz
Yes, the only way that you can get a sequence which doesn't converge to a repeating cycle is if you find a sequence with unbounded elements, but it is easy to show that this is not the case. If you st …
- 84.6k
4
votes
Accepted
Finite set of (perfect power) polynomial values?
As Qiaochu said in the comments, you must include Pell type equations as a special case, because they are the only counter example. At least for $k=2$, Siegel's theorem on integral points on algebraic …
- 84.6k
5
votes
Accepted
A parametrization of Heronian triangles
Let your triangle $\triangle{ABC}$ have side lengths $a,b,c \in \mathbb{Q}$ and rational area. Assume WLOG that $c$ is the longest side and drop the altitude from $C$ with length $h\in Q$. The triangl …
- 84.6k
14
votes
Integers n for which n divides $a^n-b^n$
The question for $a=10$ was investigated in
C. Cooper & R. E. Kennedy, "Niven Repunits and 10^n = 1 (mod n)" in 'The Fibonacci Quarterly' pp. 139-143 vol 27.2 May 1989
C. Smyth has a paper "The …
- 84.6k
8
votes
Accepted
Trying to solve: Show that n does not divide 3^n - 2^n for n greater than or equal to 2.
Take $p$ to be the smallest prime divisor of $n$. You have that $p$ divides $3^{p-1}-2^{p-1}$ and also $3^n-2^n$. So $p$ divides $3^{\operatorname{gcd}(p-1,n)}-2^{\operatorname{gcd}(p-1,n)}$. However …
- 84.6k
9
votes
Accepted
An inequality relating the factorial to the primorial.
This answer is just to point out that the result is true for large enough $n$. Let's rewrite it as $$\prod_{n\le p\le 2n}p > \sqrt{\binom{2n}{n}}$$
Since $\binom{2n}{n}\approx \frac{4^n}{n}$ introduci …
- 84.6k
2
votes
Sequence of Diophantine Equations
Have you tried looking at the density of possible $z$'s?
I think the answer might be "no". Here's my heuristics (hoping it's not bogus): the smallest $z$ we can achieve using $x$ is at least $\frac{x …
- 84.6k
8
votes
Accepted
A product of gamma values over the numbers coprime to n.
Denote $$f(n)=\prod_{k=1}^{n-1}\Gamma \left(\frac{k}{n}\right)$$ and $$ F(n)=\prod_{1\le k\le n-1, k\perp n}\Gamma \left(\frac{k}{n}\right)$$
We have $f(n)=\prod_{d|n}F(n)$ and therefore by Mobius in …
- 84.6k
24
votes
Are there ever three perfect powers between consecutive squares?
If you let $a_1,a_2,\dots$ represent the sequence $1,4,8,9,\dots$ of perfect powers, it is a conjecture of Erdos that $$a_{n+1}-a_n >c'n^c.$$
The weaker conjecture $$\liminf a_{n+1}-a_n =\infty$$ is k …
- 84.6k
14
votes
Nonzero digits in n!
I believe the best current lower bound on this is the one given by F. Luca in "The Number of Non-Zero Digits of n!" Canad. Math. Bull. 45(2002), 115-118. It is proven there that the number of non-zero …
- 84.6k
11
votes
Accepted
Numerical evidence and argument against Littlewood conjecture
In order to prove the Littlewood conjecture it is enough to prove that
$$\lim_{k\to \infty} f(n_k,\alpha,\beta)=0$$
for some subsequence $n_k$. In your case, $\alpha=\beta=\frac{\phi}{2}$. And you can …
- 84.6k
4
votes
Accepted
Questions about a product of trinomials
Start by noticing that the generating function of pentagonal numbers when working in $\mathbb Z/2\mathbb Z$ is given by
$$p(x)=1+\sum_{k=1}^{\infty}\left(x^{\frac{k(3k-1)}{2}}+x^{\frac{k(3k+1)}{2}}\ri …
- 84.6k