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Results tagged with nt.number-theory
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user 48142
Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
8
votes
$2^n$-1 consisting only of small factors
It is true that if $N > 60$, then $2^{N} - 1$ has a prime factor $> 2500$.
Here's another approach. First, observe that every prime factor of $2^{p} - 1$ is $\equiv 1 \pmod{p}$. Combining this with t …
- 19.9k
3
votes
Accepted
When are the powers of 2 sum-free mod n?
This question is very similar to the one here, and the heuristic should apply equally well. In particular, $A$ is sum-free if and only if there does not exist a $k$ with $k \ne \frac{n+1}{2}$ so that …
- 19.9k
7
votes
Accepted
primitive prime divisor of $2^{8n+4} - 1 $
No. We have that $p = 709$ is a primitive prime divisor of $2^{708} - 1$. However, $\frac{2^{708} - 1}{2^{177} + 1}$ is a multiple of the prime $q = 5521693$ and therefore $q-1 | \gamma\left(\frac{2^{ …
- 19.9k
5
votes
Accepted
$p$-th root of non-torsion points on elliptic curves
No. Lemma 3.7 on page 11 from my paper here implies that if $\mathcal{T}_{1} = {\rm Gal}(K(E[p])/K)$ and there is a normal subgroup $H \unlhd \mathcal{T}_{1}$ with order coprime to $p$ for which $E[p] …
- 19.9k
16
votes
Diophantine equation $3(a^4+a^2b^2+b^4)+(c^4+c^2d^2+d^4)=3(a^2+b^2)(c^2+d^2)$
The equation you specify defines a surface $X$ in $\mathbb{P}^{3}$, and this surface is a K3 surface. It is conjectured that if $X$ is a K3 surface, there is a field extension $K/\mathbb{Q}$ over whic …
- 19.9k
9
votes
Accepted
Congruences among primes modulo which a given polynomial has roots
Here's a survey of the possible things that can happen. In regards to your first question, given any polynomial $f(x)$, there is a positive integer $M$ so that if $\gcd(b,M) = 1$, then there are infin …
- 19.9k
9
votes
Accepted
Sets of squares representing all squares up to $n^2$
We can have the size of $S$ as small as $c \ln(n)$ for some constant $c$, and we can do this in such a way that every element of $\{1, 2, \ldots, n^2 \}$ can be represented by adding or subtracting at …
- 19.9k
4
votes
Relations of eisenstein series with eta quotient
The answer is probably some fairly basic linear algebra. For the first one, each term on the right hand side is a modular form of weight $4$ and level $2$. The space $M_{4}(\Gamma_{0}(2))$ has dimensi …
- 19.9k
8
votes
Accepted
Is $p$ is square modulo $F_p$ when $p=4k+1 > 5$?
The answer to the first question is yes, although the argument I give below is not along the lines that you were originally thinking. I will show that $p$ is a square modulo $q$ for every prime factor …
- 19.9k
40
votes
Accepted
When $\frac{a^2}{b+c}＋\frac{b^2}{a+c}＋\frac{c^2}{a+b}$ is integer and $a,b,c$ are coprime na...
Yes, there is another solution. The next one I found is a bit big, namely
$$ a = 15349474555424019, b = 35633837601183731, c = 105699057106239769. $$
This solution also satisfies the property that
$$ …
- 19.9k
12
votes
Accepted
Extension of $\mathbb Q$ which splits only at primes in $S$
For many choices of $R$ and $S$ the answer is obviously no. For example, if $R$ is empty, then the answer is no, because there are no unramified extensions of $\mathbb{Q}$.
For a more interesting exa …
- 19.9k
4
votes
Accepted
Solutions to diophantine equation
I probably put a little bit too much effort into this. The only rational point on this curve is $(0,0)$ (as well as the points at infinity $(1 : 5 : 0)$ and $(1 : -5 : 0)$).
There's a slightly non-ob …
- 19.9k
10
votes
Accepted
Cusp forms with integer Fourier-coefficients
No. Take $k = 24$ and $N = 1$. Then $\Delta^{2} = q^{2} - 48q^{3} + 1080q^{4} + \cdots \in S_{K}(\Gamma_{1}(N),\mathbb{Z})$. However, if we write $\Delta^{2} = c_{1} f_{1} + c_{2} f_{2}$, where $f_{1} …
- 19.9k
5
votes
Accepted
Representation of integers by positive definite ternary quadratic polynomials with linear terms
Yes, it is possible to extend these methods, although the picture is somewhat
less clear than in the quadratic form case. When one discusses representations by a quadratic polynomial, this is equival …
- 19.9k
6
votes
Accepted
How does this sequence grow
The answer is yes, and the number of solutions with a prime $p$ is $\lfloor \frac{p+5}{8} \rfloor$ when $p \not\equiv 1 \pmod{8}$ and is $\lfloor \frac{p+5}{8} \rfloor + 1$ when $p \equiv 1 \pmod{8}$. …
- 19.9k