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Results tagged with nt.number-theory
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user 14830
Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
10
votes
Accepted
A property of Mersenne primes
The critical points of $f$ are at $x=\pm i$, so we try a projective
(a.k.a. fractional linear) change of variable that puts these
critical points at $0$ and $\infty$, namely $x = \alpha(y)$ where
$\a …
14
votes
Fifth powers modulo a prime
[Edited to describe triple and higher-order coincidences for prime $k$, recovering the observed $0.672$ proportion for $k=5$]
Darij's pretty argument, extended by GH, nicely answers the question for …
21
votes
A quadratic form represents all primes except for the primes 2 and 11.
As GH suggests, here the relevant Eisenstein and cusp spaces
are small enough that everything can be done explicitly.
It's even a bit better than the dimensions $5+4$ suggest,
because our quadratic fo …
3
votes
Accepted
equality of two numbers which are odd powers of 2 and satisfy a certain condition
No. $(2n+1,2m+1) = (17,51)$ is a counterexample because
$$
2^{51} - 1 = 7 \cdot 103 \cdot 2143 \cdot 11119 \cdot 131071
$$
and $131071 = 2^{17}-1$ is prime. The only other counterexample with $m \le …
18
votes
Accepted
Equation $x=\phi(x)+\phi(x+1)-1$
This is OEIS Sequence A067798.
Nothing else seems to be known about it; at any rate OEIS gives no
references to the literature, only a link to
a list of further such $x$
from Giovanni Resta that ex …
16
votes
Accepted
Elliptic curves over QQ with identical 13-isogeny
[Edited mostly to include the second example, corresponding to
$(t,X) = (3,-115/126)$]
Thanks to Jordan Ellenberg for
calling attention to this nice question on his blog.
I didn't remember an exampl …
9
votes
Accepted
Elements of unit modulus in ring generated by root of unity
A.Quas already noted in his comment that ${\bf Z}[\omega_l]$ might contain a $2l$-th root of unity (he gave $l=3$ but even $l=1$ works...). But that's the only possibility: we show that the only alge …
12
votes
Fermat-like equation $c^n=a^{2n}+a^n b^n + b^{2n}$
Yes, it's a nice question. It does seem that there are no solutions in
positive integers to A. Balan's equation, and indeed no integer
solutions at all other than those with $a=0$, $b=0$, or (when $n …
13
votes
Near points in several arithmetic progressions
This is actually not true. For a counterexample, take $(a_1,a_2,a_3) = (1,\phantom. \pi,\phantom.\pi/(\pi+1))$ and $(c_1,c_2,c_3) = (0,0,1)$, so our sequences are
$\lbrace n_1 \rbrace$,
$\lbrace n_2 …
6
votes
How $a+b$ can grow when $a!b! \mid n!$
FWIW here are the first examples of $a!b! \mid n!$ with $a+b-n = v$
for each $v=1,2,3,\ldots,14$:
1 [1, 1, 1]
2 [3, 5, 6]
3 [6, 7, 10]
4 [11, 29, 36]
5 [14, 47, 56]
6 [47, 59, 100]
7 [59, 110, …
8
votes
What is the max of $n$ such that $\sum_{i=1}^n\frac{1}{a_i}=1$ where $2\le a_1\lt a_2\lt \cd...
Answered
on mathstackexchange:
the upper bound of $99$ turns out to be small enough for complete enumeration
by dynamical programming after accounting for small primes and prime powers;
the maximum …
8
votes
Accepted
A question on how polynomials split over $\mathbb{F}_p$ for large primes $p$
If $r \ll p^{1/d}$ and $p \mid f(r)$ then (since $f(r) \neq 0$)
$f(r) = ap$ for some nonzero $a \ll 1$. Hence for each of
finitely many choices of $a$ we are asking for prime values of $f(r)/a$
as $r …
5
votes
Consecutive rising sequence of largest prime factors
As Kevin Buzzard suggests in a comment,
this would be a consequence of one of the
"standard conjectures on primes", namely the
first
Hardy-Littlewood conjecture
(which is the special case of
Schinze …
16
votes
What is the rational rank of the elliptic curve x^3 + y^3 = 2?
To add to Abhinav's answer: the fact that $x^3+y^3=2$ has no solutions
other then $x=y=1$ is attributed by Dickson to Euler himself:
see Dickson's History of the Theory of Numbers (1920) Vol.II, Chap …
21
votes
Number fields with same discriminant and regulator?
Building on G.Myerson's answer and KConrad's explanation, it's not hard to construct pairs $K,K'$ of quartic fields that have both the same discriminant and the same regulator. [Edited to add example …