31,125 reputation
798161
bio website math.harvard.edu/~elkies
location Harvard University, Cambridge, MA 02138
age 48
visits member for 3 years, 4 months
seen 1 min ago
[Not sure yet what to use this space for...]

14h
comment $L^2$ discrepancy bound for sequences in $[0,1)$
But the context of Schmidt's discrepancy theorem indicates that the question is what happens not for typical sequences but for any sequence, no matter how well distributed. Since there are sequences whose $L^\infty$ discrepancy is $O(\log n)$, the same sequence has $L^2$ discrepancy at worst $O(\sqrt{\log n})$. Conceivably this can be improved further even though for $L^\infty$ it is known that $C\log n$ is best possible.
Aug
30
comment Realizing algebraic curves as complete intersections
You write "$g=1,\ldots,5,9,10,12,16,\ldots$", but $g=2$ does not occur (while $g=0$ of course does).
Aug
29
awarded  Enlightened
Aug
29
awarded  Nice Answer
Aug
29
revised Approximating rational values in ]0,1[ by a sum or difference of unit fractions
(u_1, u_2), not (u_1, u_1) ...
Aug
29
comment Elliptic curves with trace of Frobenius values always congruent to 0 modulo 2
The fact that $\omega_2 = 1$ is immediate because the determinant is identically $1$ on ${\rm GL}_2({\bf Z}/2{\bf Z})$...
Aug
28
answered Approximating rational values in ]0,1[ by a sum or difference of unit fractions
Aug
28
answered Elliptic curves with trace of Frobenius values always congruent to 0 modulo 2
Aug
25
comment Finding integer points on elliptic curves via divisibility conditions like $(a+b)^2 \mid (2b^3+6ab^2-1)$
[The condition if(l%3, (i.e., don't try $\ell$ if $3\mid\ell$) exploits the fact that $x^3-2$ has no roots mod $3^2$. I could have also saved a factor of nearly $3/4$ by requiring that $\ell$ not be divisible by $7$, $13$, or $19$.]
Aug
25
comment Finding integer points on elliptic curves via divisibility conditions like $(a+b)^2 \mid (2b^3+6ab^2-1)$
OK, gp code follows, though I'm afraid the line breaks and indentations will be lost $-$ or is there a way to retain such formatting in comments? $$ $$ L = 10^8; { forstep(l=3,L,2, if(l%10^6==1,print("<",l-1,">")); if(l%3, F = factor(l); n = #F[,1]; v = vector(n,i,polrootspadic(x^3-2,F[i,1],2*F[i,2])); for(i=1,n, v[i] = lift(v[i]) * Mod(1,F[i,1]^(2*F[i,2]))); forvec(r=vector(n,i,[1,#v[i]]), m = Mod(0,1); for(i=1,n, m=chinese(m,v[i][r[i]])); m = lift(m); if(m<l,print([l,m])) ) ) ) }
Aug
25
comment Finding integer points on elliptic curves via divisibility conditions like $(a+b)^2 \mid (2b^3+6ab^2-1)$
It's faster to look for $\ell^2 \mid m^3 - 2$ via the factorization of $\ell$ rather than $m^3-2$. A 30 minute calulation in gp (using polrootspadic) finds that there are no examples of $m<\ell$ with $\ell$ odd and $0 < \ell < 10^8$ besides the known $(\ell,m) = (5,3)$ and $(127,100)$.
Aug
22
comment An interesting calculation of derivative
You're welcome, and thank you too!
Aug
22
answered An interesting calculation of derivative
Aug
21
comment Upper bound on the number of ismorphism classes of bilinear forms on $\mathbb{Z}^n$
An upper bound can be obtained by multiplying the mass formula by an upper bound on the number of automorphisms of the form. That bound grows quickly with $n$, but not nearly as fast as the mass formula, so the resulting upper bound should be reasonably good (at least its logarithm will be of the right size). For large $n$ it's probably true that most forms have automorphisms only by $\{ \pm 1 \}$, which would mean that the mass-formula bound is asymptotically sharp. I don't know whether this has been proved, or how hard it would be to prove it.
Aug
14
comment When is a cubic polynomial a cube?
You probably know already that $P_n(x) := ax^3+bx^2+cx+d$ is identically a cube if $n=0$ or $\pm1$. Otherwise $P_n(x)$ has no repeated factors so $y^3=P_n(x)$ is an elliptic curve and thus has only finitely many integer solutions, but it can be hard to provably list them all and we don't expect to be abls to do it uniformly in $n$. I suppose you know already that $x=(1-n)/2$ always works if $n$ is odd, while both $x=-n/2$ and $x=1-(n/2)$ work if $n$ is even. There are also occasional sporadic solutions like $x=-34,-22,3,15$ for $n=20$.
Aug
10
revised Szemeredi's theorem in the Gaussian integers
Add Tao/Ruzsa/Freiman link (and correct misspelling of "Szemeredi")
Aug
10
awarded  Enlightened
Aug
10
awarded  Nice Answer
Aug
10
revised Szemeredi's theorem in the Gaussian integers
Correct 3AP equation: a1+a3, not a3+a3...
Aug
10
answered Szemeredi's theorem in the Gaussian integers