bio  website  math.harvard.edu/~elkies 

location  Harvard University, Cambridge, MA 02138  
age  48  
visits  member for  4 years, 3 months 
seen  40 mins ago  
stats  profile views  24,510 
[Not sure yet what to use this space for...]
1d

comment 
Are there infinitely many $k$ for which $\frac{\sigma(k)}{k}=n^p$ and $p$ is an odd prime?
BTW there should be infinitely many $k$, such as $51590^2$, for which both $k$ and $\sigma(k)$ are square, and thus $\sigma(k)/k$ is a rational square. Perhaps this already follows from known results. 
2d

comment 
Is there a way to find an efficient set of relations for presenting the subgroup generated by two matrices in $SL(2, q)$?
Use the classification of subgroups of $\text{SL}_2$. The hardest case is finding generators and relations for $\text{SL}_2(\mathbb F_q)$ itself, and that must be known. 
Jul 27 
comment 
Recent progress on the busy beaver problem?
Or at least lower bounds on it (the last known value is already so large that we probably won't ever see more than one or two further values). 
Jul 23 
comment 
Intuition behind the Kodaira Vanishing Theorem?
If you want an intuition for a theorem about "$H^q$" of a "line bundle with positivedefinite curvature form", you probably need an intuition for what $H^q$ and that condition on $L$ mean. Is there a standard intuitive/heuristic interpretation of these notions? 
Jul 23 
comment 
How can I solve a diophantine equation with 3 variables?
Well, questions of implementation are not automatically offlimits here; very similar questions can lead to nontrivial issues in integer programming. But in this case the parameters are too small (and too specific without evident motivation); FWIW the mindless gp code forvec(xyz=vector(3,dummy,[21,99]),if(xyz*[1,17,289]~==3367,print(xyz))) comes up empty. 
Jul 22 
comment 
Integral points on a particular family of curves
Maybe easier for $n=6$, because there the curve is bielliptic [involution $(x,y) \leftrightarrow (7x,y)$] and it's easier to compute ranks of two elliptic curves than one genus2 Jacobian. There's also an elliptic quotient for $n=8$. 
Jul 20 
comment 
Transformations that leave the Plucker embedding of G(2,4) invariant
PGL and PSL are the same over C, though. (Multiply the transformation of determinant $1$ by $i$ or any other sixth root of $1$; the resulting transformation does not quite preserve the quadric form, but does preserve its zerolocus.) 
Jul 20 
comment 
Transformations that leave the Plucker embedding of G(2,4) invariant
It's actually twice as big, right? There are automorphisms of determinant $1$ that come from an outer automorphism of PSL(4). 
Jul 20 
comment 
Transformations that leave the Plucker embedding of G(2,4) invariant
Over the complex numbers there's no signature; it's just PGO(6). 
Jul 17 
comment 
Determining when combinatorial sums are zero
This should settle the vanishing question, at any rate: according to this paper math.berkeley.edu/~lam/html/fila.ps Schur showed in 1929 that the Laguerre polynomials are always irreducible, so in particular cannot have a rational root once $n>1$. 
Jul 7 
awarded  Good Answer 
Jul 7 
comment 
Are these two $q$continued fractions equivalent?
Presumably algebraic numbers only at special values of $q$ (namely $e^{2\pi i z}$ where $z$ is a quadratic irrationality in the upper halfplane; e.g. $z=i$ gives your $e^{2\pi}$). 
Jul 7 
comment 
Why the letter “p” for genus?
Wiktionary gives several senses for род including "genus". I tried to give the link but it doesn't seem to follow; here it is in two pieces: en.wiktionary.org/wiki followed by /род 
Jul 7 
revised 
Minimum value of $p(1)^2+p(2)^2 +…+ p(n+3)^2$ over all monic polynomials $p$
explain a few steps and connect with the Hahn polynomials 
Jul 6 
awarded  Enlightened 
Jul 6 
awarded  Nice Answer 
Jul 6 
comment 
How small can the MumfordTate group of hypersurface be?
You probably want $d>4$, because it's still dense in ${\cal A}_g$ for $d=4$. 
Jul 6 
answered  Minimum value of $p(1)^2+p(2)^2 +…+ p(n+3)^2$ over all monic polynomials $p$ 
Jul 4 
comment 
Minimum value of $p(1)^2+p(2)^2 +…+ p(n+3)^2$ over all monic polynomials $p$
Well $n$ or fewer is trivial :) and I guess you're expected to see how to do $n+1$ before figuring out how to deal with a few more values. 
Jul 4 
comment 
Existence of Hecke operators with distinct eigenvalues?
To construct a counterexample, use three forms $f_1,f_2,f_3$ in the same space (they can even be weight2 newforms) that are quadratic twists of each other. For each $p$ at least two of the $f_i$ must have the same $T_p$ eigenvalue. 