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Results tagged with nt.number-theory
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user 7076
Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
1
vote
The existence of solutions of a system of indeterminate equations
I expect that in many cases the following construction will do the job. By Fermat polygonal number theorem, we have representation of $\frac{m(m-3)}4$ as the sum of 3 triangular numbers:
$$ \frac{m(m- …
- 29.5k
2
votes
Accepted
The existence of solutions of a system of indeterminate equations
In the comments OP proposed a greedy algorithm to represent a given positive integer $A$ as the sum of triangular numbers whose indices sum to $m$, and applied it to $A = \frac{m(m+1)}4$. I will prove …
- 29.5k
0
votes
Accepted
Is there a general way to solve this modular equation?
There seems to be no simple formula for the solutions to the given congruence. Still, they can be computed iteratively as follows.
First, we notice that $2^m = (3-1)^m \equiv (-1)^m \sum_{i=0}^{N-1} \ …
- 29.5k
2
votes
Are there infinitely many composite $a$ such that $\sum_{k=1}^{a}(k,a)\equiv1\pmod{a-1}$?
I do not know answer to your question, but here is an approach how one can extend a given integer $m$ to a solution $mp$ (if one exists) with a prime $p\nmid m$.
Using the multiplicativity, we want
$$ …
- 29.5k
2
votes
Accepted
A problem on generators and Hensel lifting
Write $g'=g(1+ap)$ so that
$$g'^y\equiv g^y(1 + pya)\pmod{p^2}.$$
It follows that we can take
$$a = \frac{(\ell^2/g^{y}\bmod p^2)-1}{py},$$
which can be computed in polynomial time.
- 29.5k
9
votes
$23005\cdot (2^n-1)\cdot 2^n +1=p^2$
Multiplying the equation by $2^n$ and denoting $X:=2^n$ and $Y:=p2^{\lfloor n/2\rfloor}$, we obtain two elliptic curves (depending on the parity of $n$):
$$(23005(X-1)X+1)X=Y^2,$$
$$(23005(X-1)X+1)X=2 …
- 29.5k
13
votes
Accepted
Integers $b$ such that $n \nmid (b^n-1)$ for $n>1$
$b=2$ is the only almost 2-like number. Indeed, if $n\mid (b^n-1)$ and $p$ is a prime divisor of $(b^n-1)/n$, then $np\mid (b^{np}-1)$. That is, existence of one $n>1$ dividing $b^n-1$ implies existen …
- 29.5k
6
votes
Mid-Square with all bits set
I'd like to point out why the case of middle bits being all ones is somewhat special and different from other fixed values of them. Also, there is an approach for finding a suitable numbers that may n …
- 29.5k
3
votes
Accepted
Runs of consecutive numbers all of which are Murthy numbers
Let $m+1, \dots, m+n$ be a sequence of $n$ consecutive Murthy numbers such that each $m+i$ shares with its reversal $\overline{m+i}$ a prime factor $p_i\equiv 3\pmod4$ such that 10 is a quadratic nonr …
- 29.5k
3
votes
What is the highest power of 2 that divides $3^y(2z-1)-1$?
I doubt there is a simple expression for $x$ as a function of $y,z$. However, it is possible to characterize all cases when $2^k\mid 3^y (2z-1) - 1$ for a given positive integer $k$.
Say, for $k\geq 3 …
- 29.5k
1
vote
A variant of Landau's function
Let $k = \nu_2(g(n))$. Then $g(n)$ is attained at $n = 2^k + \texttt{1s and powers of odd primes}$, implying that $g(n) = h(n-2^k)2^k$.
Hence, $h(n) = \frac{g(n+2^k)}{2^k}$ for some $k$ satisfying $k= …
- 29.5k
2
votes
Accepted
Solutions of a exponential diophantine equation involving the $\sigma$ function
Note that
$$z = \frac{p^{4k+1}}{\sigma(p^{4k+1})} = \frac{1-p^{-1}}{1-p^{-(4k+2)}} > 1-p^{-1}.$$
Clearly, for a fixed p, the larger is $k$ the closer is $z$ to this lower bound.
Since the lower bound …
- 29.5k
12
votes
Is $441$ the only square of the form $\frac{397\cdot 10^n-1}{9}$?
If $\frac{397\cdot 10^n - 1}9$ is a square then so is $397\cdot 10^n - 1$. Let $y^2=397\cdot 10^n - 1$. Denoting $x:=10^{\lfloor n/3\rfloor}$, we get that
$$y^2 = 397\cdot 10^r\cdot x^3 - 1$$
or
$$(39 …
- 29.5k
7
votes
Accepted
A modification of the Ljunggren-Nagell equation
The equation
$$
\frac{a^m-1}{a-1}=2b^2
$$
does not have solutions in positive integers for $m>2$ as shown below.
First, notice that $a$ must be odd.
Second, one can see that $m$ must be even. Indee …
- 29.5k
2
votes
Simultaneous lcms
For each prime $p|d$, let $q_p$ be the number of $n_i$ with $p|n_i$.
Then the number of ordered but not necessarily distinct solutions $(m_1,\dots,m_r)$ is given by
$$f(r)=\prod_{p|d} S(r,q_p)\cdot q_ …
- 29.5k