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location Madrid, Spain
age 51
visits member for 3 years, 11 months
seen 16 hours ago

Apr
2
comment Best Fake Proofs? (A M.SE April Fools Day collection)
It seems that Euler himself was confused with this. See [here]( webspace.utexas.edu/aam829/1/m/Euler_files/EulerMonthly.pdf)
Mar
9
awarded  Guru
Jan
19
answered $n +1$th Fibonacci number modulo $n$
Jan
4
comment Exact power of $p$ that divides the discriminant of an algebraic number field
The statement will be easier to follow if you suffix the $\beta$'s. If I understand, there are $f_i$ elements in $B_i$, say $\beta_{ik}$ for $1\le k \le f_i$ and then the $n$ elements $\alpha_1,\dots,\alpha_n$ are $\alpha_{ij}\beta_{ik}$ for $1\le i\le r$, $1\le j\le e_i$ and $1\le k \le f_i$. In addition I think the $x \equiv \alpha_{ij} \pmod{Q_i^{e_i}}$ should read $x \equiv 1 \pmod{Q_i^{e_i}}$. If it is so, then the "large number $N$" means possibly: $N \ge e_i$ for every $i$ as you are then free to chose $\alpha_{ij}$ with the stronger constraint $\alpha_{ij} \equiv 0 \pmod{Q_h^N}$.
Dec
1
revised Galois group and the Quaternion group
added 1 characters in body
Dec
1
comment Galois group and the Quaternion group
I have corrected the right hand side of $(\theta^2-6-2\sqrt{3})^2$, sorry for that. I have also added some computations in the end to help you with the automorphisms.
Dec
1
revised Galois group and the Quaternion group
Added explanation
Nov
30
answered Galois group and the Quaternion group
Nov
10
comment Early history of lower bounds on the prime counting function
I'm sorry I've made an incorrect statement, I could not edit it so I have removed it.
Nov
8
awarded  Yearling
Nov
7
answered Cover a disk with thin rectangles
Nov
6
comment What would be complexity of computing $3^{n^n}$?
If you are computing $3^{n^n}$ modulo an integer $M$ and you know it's totient function $\varphi(M)$ then you can compute $n^n$ in $O(\log n)$ multiplications $\pmod{ \varphi(M)}$ and then rise $3$ to the result in $O(\log M)$ multiplications mod $M$ finding the result in polynomial time. The problem is to find $\varphi(M)$. Some times it can be done in polynomial time (for example if $M$ is prime then $\phi(M)=M-1$), but in general if $M$ is composite finding $\varphi(M)$ is equivalent to finding the factorization of $M$.
Nov
5
comment Proving a simple inequality
I find $$f(3) = \frac{\tfrac{2}{3}\log 2}{\log 6} = 0.2579... > f(4) = \frac{\tfrac{1}{2}\log 3}{\log 12} = 0.2210...$$ Are you sure the inequality is correctly written?
Nov
3
revised Half the rationals?
Added a generalization
Nov
3
answered Half the rationals?
Nov
1
comment Half the rationals?
A good candidate for your set $X$ with a very simple description is the set of reduced fractions $a/b$ with $$ a\cdot b \equiv 0 \pmod{3} $$
Nov
1
comment Half the rationals?
I think there something missing in the right hand side of the inequality $\sum_{k=1}^{n-1} a_k\phi(k) < 1/2$, I suppose you mean $<\frac{1}{2}\sum_{k=1}^{n-1} \phi(k)$? On the other hand, even if it is probably true, I can't see how to prove that this construction works for all the intervals $Y$.
Nov
1
comment Half the rationals?
I think that the proportion of fractions with even numerator and denominator bounded by $N$ is aproximately one third, and the same for even denominator.
Oct
15
revised Tiling a minimal perimeter region with $n$ unit squares
Sketch of proof
Oct
15
answered Tiling a minimal perimeter region with $n$ unit squares