35,825 reputation
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bio website math.harvard.edu/~elkies
location Harvard University, Cambridge, MA 02138
age 48
visits member for 3 years, 11 months
seen 7 hours ago
[Not sure yet what to use this space for...]

Apr
12
comment Models for the moduli space $\overline{M}_{1,n}$
I thought it was only rank $9$ (with further tricks to find rank $10$ and $11$ even though $M_{1,11}$ and $M_{1,12}$ are not rational). So it's $T_1$ through $T_{18}$, not $T_{20}$; and moreover the first $8$ parameters can be removed thanks to the ${\rm PGL}_3$ automorphisms of the projective plane, leaving a rational parametrization of the moduli space $\overline{M}_{1,10}$ by $T_9,\ldots,T_{18}$.
Apr
12
comment Models for the moduli space $\overline{M}_{1,n}$
I think it's rational only for $n \leq 10$ and is birationally parametrized by configurations of $n-1$ points in the plane and the cubic(s) through them. (The last point is obtained using the divisor $O(1)$ on the cubic.)
Apr
4
comment Diophantine equations and the numbers $4,7,8$
If $x,y,z$ are positive then $x^n + y^n + z^n \geq 3 (xyz)^{n/3}$ (AM-GM inequality), so once $n>3$ it's a finite search.
Apr
3
comment Finding extrema of a cubic equation
1) The problem has been concocted to have a rational solution. Finding rational roots of a cubic (e.g. by factoring) is fair game. Or 2) divide by $2$ to get $216/(10-x)^3 = 125/x^3$ and extract cube roots.
Mar
28
comment Are there any Algebraic Geometry Theorems that were proved using Combinatorics?
(A generic smooth hypersurface, I suppose you mean: there are certainly hypersurfaces of each degree and dimension with nontrivial automorphism groups.)
Mar
22
comment About distinct eigenvalues of a graph
Wait, a disconnected graph does not satisfy this condition for any polynomial $p$, and can easily have all eigenvalues distinct (e.g. the graph with 3 vertices and 1 edge). [Presumably $A$ is the adjacency matrix, though the problem statement never actually defines $A$.]
Mar
22
comment Enumeration of $0-1$ matrices with determinant $1$
Why is that a reasonable guess? There are $2^{n^2}$ zero-one matrices, each with $|{\rm disc}| < n^{n/2}$, so I'd expect $2^{n^2 - O(n\log n)}$.
Mar
20
comment Can you find squares in this class?
@Michael Stoll For $p=997$, your ratpoints program doesn't take long to find the minimal solution $(l,m) = (130792, 148329)$.
Mar
18
comment Dynamics of electrons on a sphere
The answer to Q1 must be Yes, because the differential equation has a unique solution so it must retain the initial symmetry. For Q2 the particles must approach a stable local minimum, but in general they don't have to find a global minimum, and the probability of success might depend also on the strength of the damping. (Is a regular $n$-gon on the equator locally stable once $n>3$?)
Mar
15
revised Structure of sign changes under the heat flow
Explain why $u$ solves the heat equation.
Mar
14
awarded  Enlightened
Mar
14
awarded  Nice Answer
Mar
14
awarded  Necromancer
Mar
14
awarded  Revival
Mar
14
answered Degree 17 number fields ramified only at 2
Mar
12
revised Structure of sign changes under the heat flow
Change notation from $N_u(\ldots)$ to $V_u(\ldots)$ to conform with OP's notation
Mar
12
answered Structure of sign changes under the heat flow
Mar
12
comment The Maximal $\ell_2$ norm of a signed sum of vectors
There are $2^{n-1}$ possibilities, so for $n=3$ it's just the maximum of four candidates, which is easy to compute. For example, let $G$ be the Gram matrix with $(i,j)$ entry $G_{ij} = v_i \cdot v_j$. Then the maximum norm is at most $\|G\|^{1/2}$ where $\|G\| := \sum_{i,j=1}^3 |G_{ij}|$. Equality holds unless all $G_{i,j}$ entries are nonzero and an odd number of the entries above the diagonal are negative, in which case the maximum is the square root of $\|G\| - 2 \min_{i,j} |G_{ij}|$.
Mar
11
comment $x^4+y^4$ powerful for relatively prime $x,y$
Yes I checked, but not that way: you don't want to wait for gp to count to $10^{16}$, let alone factor every number of at most $16$ digits! Much better to try all coprime $(x,y)$ of opposite parity with $x<y$ and $x^4 + y^4 < 10^{16}$; that's only 40 million or so factorizations, which take a few hours to try (and as expected find nothing). Still it's hopeless to reach $3 \cdot 10^{36}$ this way... Now that it's a couple of months since I posted this question, I should post a partial answer evaluating different strategies, the best of which might make the computation barely feasible.
Mar
11
comment $2^n$-1 consisting only of small factors
According to that Wikipedia article, the result on $2^n-1$ is actually an earlier theorem of Bang which Zsigmondy generalized.