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  
stats  profile views  20,201 
[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 $(a+b)^2 \mid (2b^3+6ab^21)$
[The condition if(l%3, (i.e., don't try $\ell$ if $3\mid\ell$) exploits the fact that $x^32$ 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 $(a+b)^2 \mid (2b^3+6ab^21)$
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("<",l1,">")); if(l%3, F = factor(l); n = #F[,1]; v = vector(n,i,polrootspadic(x^32,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 $(a+b)^2 \mid (2b^3+6ab^21)$
It's faster to look for $\ell^2 \mid m^3  2$ via the factorization of $\ell$ rather than $m^32$. 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 massformula 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 $x=(1n)/2$ 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 