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Q&A for professional mathematicians
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Hottest Questions Today - MathOverflow
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I hope everyone is doing well. Let $K \subset \mathbb{R}^n$ be a centrally symmetric convex body $(K = -K)$. Denote by $K \mid H$ the orthogonal projection of $K$ onto $H$, where $H$ is an $n - 1$
Mathoverflow.net news digest
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0 days
Rational points on explicit genus-one quartics from a finite descent computation
I have been working on a finite packet descent related to the perfect cuboid problem. The reduction is now down to four surviving packets, each reduced to explicit genus-one curves. I am not asking for a proof of the perfect cuboid problem here; I am...
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5 days
On $a^3+b^3+c^3 = N$ and $a^5+b^5+c^5+d^5+e^5 = N$
(Update, June 27, 2026): Jeff Braun found one of the missing $N\equiv 5 \pmod{11}$,
$$49 = - 22403^5 + 21596^5 + 15669^5 + 4698^5 -4001^5$$
using a search radius $|x_i|<40000$. -
0 days
I computationally have found the following conjecture to hold true for $1 \le k \le 12$, $4 \le m \le 15$:
Conjecture. For integers $k \ge 1$, $m \ge 4$, $n \ge 1$, let $x_k(m) = \bigl\lfloor m(2^m+1)/(2k)\bigr\rfloor + 1$. Then for all $n \ge x_k(m)$,$$0 \;\le\; \frac{(kn)^m - 2^m}{2} - \frac{1}{\displaystyle\sum_{j=1}^{kn-1}\binom{kn}{j}^{-m}} \;<\... -
1 year
Lewin dilogarithm formula for $\operatorname{Li}_2\left(r e^{\pi i / 6}\right)$
The following polylog formula is well known.
$- \operatorname{Li}_2(r) + \frac{1}{3} \operatorname{Li}_2(r^3) - 2 \operatorname{Li}_2\left(r e^{2\pi i / 3}\right)=0$ for real $r$.
Lewin investigated a more complicated version with $\operatorname{Li}_2\left(r e^{\pi i / 6}\right)$ instead....
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