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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
4
votes
Generalization of Wilson's theorem for prime tuples
I assume that all $h_i$ are even.
Then $(n+h_1,n+h_2,\dots,n+h_k)$ is a prime tuple iff
$$\begin{cases}
h_1!m \equiv -1\pmod{n+h_1},\\
\dots\\
h_k!m \equiv -1\pmod{n+h_k},
\end{cases}
$$
where $m=(n-1 …
- 29.5k
4
votes
Accepted
Primes in shifted geometric sequence
The answer is No.
We know that $78557$ is a Sierpinski number with the covering set $S = \{ 3, 5, 7, 13, 19, 37, 73 \}$, i.e., every integer of the form $78557\cdot 2^n+1$ is divisible by an element …
- 29.5k
5
votes
Accepted
On a system of equations in $\mathbb F_2$
1 - No, the problem is NP-hard even if degree of all polynomials is 2.
2 - See, for example, paper Solving Systems of Polynomial Equations over
GF(2) by a Parity-Counting Self-Reduction by Björklund e …
- 29.5k
2
votes
Accepted
Diophantine equations that involve Lehmer means with all digits equal to $1$ in their $x-$ad...
Conjecture 1 does not hold as for any even $n$, (2) has a solution:
$$(x,y,z)=(2,\frac23(2^n-1),2(2^n-1))$$
Likewise, Conjecture 2 fails as for $p=2$ and any even $n$, (1) has a solution:
$$(x,y,z)=(4 …
- 29.5k
24
votes
Accepted
Simple recurrence that fails to be integer for the first time at the 44th term
Copying my explanation from https://mathoverflow.net/a/217894/25028
The recurrence formula can be rewritten as
$$a_2=2,\qquad a_{n+1}=\frac{a_n\cdot (a_n+n-1)}n,\quad n\geq 2,$$
which somewhat justif …
- 29.5k
5
votes
Parametrization of integral solutions of $3x^2+3y^2+z^2=t^2$ and rational solutions of $3a^2...
For #1, we can take a particular solution such as $(a_0,b_0,c_0)=(0,0,1)$, and search parametric solution in the form: $(a,b,c)=(a_0+\alpha t, b_0+\beta t, c_0+t)$. Plugging it into the equation and s …
- 29.5k
1
vote
Accepted
Analytical result of a combination like generating function
The limit in question equals the coefficient of $z^X$ in $f(\varphi,z)$. For $1\leq X\leq N$, series multisection allows to express this coefficient in the following closed form:
\begin{split}
& \frac …
- 29.5k
5
votes
Accepted
$\sum_{1\leq a,b\leq n}\mathcal{P}(\gcd(a,b,n))$
Introducing $d:=\gcd(a,b,n)$, we get
$$\sum_{1\leq a,b\leq n} \mathcal{P}(\gcd(a,b,n)) = \sum_{d\mid n} \mathcal{P}(d) J_2(\tfrac{n}d),$$
where $J_2(\cdot)$ is the Jordan totient function (see also OE …
- 29.5k
-1
votes
A set, product of any two elements minus one is a perfect square
The system of Diophantine equations like
$$\begin{cases}
2d-1 = x^2\\ 5d-1 = y^2 \\ 13d-1 = z^2
\end{cases}$$
can be systematically computationally solved by considering an equivalent system (with eli …
- 29.5k
2
votes
Accepted
Finding all proper divisors of $a_3z^3 +a_2z^2 +a_1z+1$ of the form $xz+1$
Such an extension is highly unlikely to exist. Already in the simple case of $z=2$, it's equivalent to just factoring a given odd integer $n$, which is a famous hard problem.
- 29.5k
7
votes
Accepted
Reduced resultants and Bezout's identity
First off, it can be seen that the reduced resultant must divide $c$. Indeed, if $u_1(x)f(x)+v_1(x)g(x)=c$ and $u_2(x)f(x)+v_2(x)g(x)=d$, then from the Bezout identity for integers $c,d$, we can const …
- 29.5k
7
votes
For which $n$ is $\sum_{k=1}^n 1 / \varphi(k)$ an integer?
Here is some computational evidence that $n\in\{1,2,4\}$ are the only $n$ that deliver an integer sum.
For an odd prime $p$, let $a_p < b_p$ denote two smallest even positive integers such that $a_pp+ …
- 29.5k
3
votes
how to solve this symmetrical equation in number theory
It is convenient to define $n_1:=m_1-2^{\alpha-\gamma}$, $n_2:=m_2-2^{\alpha-s}$, and $n_3:=m_3-2^{\alpha-t}$. Then the equation in question takes form:
$$n_1n_2n_3 = (2^{\alpha}-1)(2^{\alpha-s-t}n_1+ …
- 29.5k
3
votes
Sieving modulo non-prime residue classes
Let $D=\{ d>1\ :\ d\mid n \}$ be the set of nontrivial divisors of $n$. Then by inclusion-exclusion, the number of elements remaining after sieving based on $a\equiv 0\pmod{d}$ equals
$$\sum_{S\subset …
- 29.5k
1
vote
Show that, If $a-1\mid S(a-1,2m)$ and $a-1>2m+1$ then $(f(a,2m))_a\in X_a$
Clearly, we have $(a-1)|S(a-1,2m)$ iff $(a-1)|D(a,S(a-1,2m))$.
Let $q:=\frac{D(a,S(a-1,2m))}{a-1}$. Since for $a\geq 4$ and $m\geq 1$,
$S(a-1,2m) < (a-1)a^{2m}$ and $S(a,2m) = S(a-1,2m) + a^{2m}$, …
- 29.5k