Esteban Crespi
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 Mar 14 revised Reducing simultaneously a pair of fractions $\frac{a^2}{b},\frac{ a^3}{c}$ using only gcds Sorry the method I give is not correct Mar 14 asked Reducing simultaneously a pair of fractions $\frac{a^2}{b},\frac{ a^3}{c}$ using only gcds Mar 1 comment Closed form for $_2F_1\left(\frac12,\frac23;\,\frac32;\,\frac{8\,\sqrt{11}\,i-5}{27}\right)$ Could you give a reference to understand how the value $g_3 = \eta^6/16$ is found? Thanks. Feb 13 answered $17$ is a quadratic residue for all primes $p$ such that $p \equiv \pm 3 \mod{8}$? Dec 22 awarded Necromancer Dec 22 revised Generalizing Ramanujan's proof of Bertrand's Postulate: Can Ramanujan's approach be used to show a prime between $4x$ and $5x$ for $x \ge 3$ minor edit Dec 22 awarded Revival Dec 22 answered Generalizing Ramanujan's proof of Bertrand's Postulate: Can Ramanujan's approach be used to show a prime between $4x$ and $5x$ for $x \ge 3$ Dec 18 awarded Nice Question Nov 8 awarded Yearling Jun 21 revised How prove that:$\varphi(2)+\varphi(3)+\varphi(4)+\cdots+\varphi(n)\ge\frac{n(n-1)}{4}+1$ I have rewritten the proof as the original was very confusing and contained some incorrect statements. Jun 21 revised How prove that:$\varphi(2)+\varphi(3)+\varphi(4)+\cdots+\varphi(n)\ge\frac{n(n-1)}{4}+1$ deleted 2 characters in body Jun 21 answered How prove that:$\varphi(2)+\varphi(3)+\varphi(4)+\cdots+\varphi(n)\ge\frac{n(n-1)}{4}+1$ Jun 4 comment Can this sum be simplified to closed form? I'm not sure how to prove it but it seems that the following might be true: set $S(L,b) = \sum_{a=0}^{L-1} \lfloor ab/L\rfloor^2$, and write $L = nb+t$ then $S(L,t) = n(n+1)(2n+1)/6 + S(t,b)$. May 4 comment Do these inequalities regarding the gamma function and factorials work? Is $x$ an integer? if not what does it mean $n \vert (x+1)$? 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