bio  website  math.brown.edu/~jhs 

location  Brown University Mathematics Department  
age  
visits  member for  4 years, 2 months 
seen  12 hours ago  
stats  profile views  5,325 
Professor of Mathematics, Brown University.
Primary interests: number theory, arithmetic geometry, elliptic curves, arithmetic dynamics, cryptography.
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awarded  Good Answer 
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comment 
Uninteresting questions with interesting answers
@Wojowu It was assigned as a homework problem when I took undergrad differential geometry class, at which time I knew who it was named after. But I'm afraid I've forgotten even that piece of information. And I don't know any references. Sorry. 
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Uninteresting questions with interesting answers
@RobinSaunders Probably it's fine for nonconvex curves with signed area, as you say. It's also okay for piecewise smooth curves. In any case, the proof is a lovely application of Green's theorem, and from the proof one should be able to see how to formulate it more generally than I did in the answer that I gave. But note, for example, if $L$ is too long and $C$ is too small, it may not be possible to slide $L$ around inside $C$, so some restrictions are necessary. 
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Uninteresting questions with interesting answers
@WillieWong Yes, $C$ is supposed to be closed. 
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awarded  Nice Answer 
Mar 24 
answered  Uninteresting questions with interesting answers 
Mar 14 
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Parodies of abstruse mathematical writing
@Vincent Yes, that's the one, but now it's "linked to above", since his answer quite rightly has more upvotes than mine! 
Mar 14 
answered  Parodies of abstruse mathematical writing 
Mar 6 
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Variety acquiring rational point over any quadratic extension
@MichaelStoll Yes. The proof (which isn't completely straightforward, but not hard) is in "Bielliptic curves and symmetric products." Proc. Amer. Math. Soc. 112 (1991). It was greatly generalized by Abramovich and Harris in "Abelian varieties and curves in $W_d(C)$." Compositio Math. 78 (1991). 
Mar 6 
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Periodic points in C^2
Write $f^n(x,y)=(P_n(x,y),Q_n(x,y))$. Then the points of period dividing $n$ are the solutions to the pair of equations $P_n(x,y)=x$ and $Q_n(x,y)=y$. Each of these equations defines a curve, and each of these curves has a finite number of irreducible components. There will be an infinite number of solutions in $\mathbb{C}^2$ if and only if these two curves have a common component. If your map extended to a regular map on $\mathbb{P}^2$, then it's easy, but your map is not defined at $[1,0,0]$. So one probably needs to do some nontrivial geometry. 
Mar 1 
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Regularized sums of Mobius sequence
I'm sure I'm missing something, but why does $\int_0^\infty x^s\,\frac{dx}{x}$ converge "a priori" for Re$(s)>1$? For, say, $s=2$, certainly $\int_0^\infty x\,dx$ doesn't converge. 
Feb 10 
answered  Unable to find any information regarding this fact (Frey, elliptic curves) 
Feb 6 
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Lower bound for spectral radius on $\operatorname{GL}(n,\mathbb{Z})$
@LiamBaker No, I didn't say that they are equivalent, I very deliberately used the words "highly likely," which means I was making a guess. So my guess is that this is the hardest case, and that a proof of that case would very likely end up being a proof of the full conjecture. But that's all speculation, and I also wouldn't be shocked to be wrong. 
Feb 6 
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Lower bound for spectral radius on $\operatorname{GL}(n,\mathbb{Z})$
@YemonChoi (Exponential) Mahler measure $M(f)$ is the product of $\lambda$ over all roots satisfying $\lambda>1$. So $M(f)\ge R$, since $R$ is just the largest root. On the other hand, it's highly likely that the smallest value of $M(f)>1$ occurs for a polynomial that has one root $\lambda$ outside the unit circle, one root $\lambda^{1}$ inside the unit circle, and all other roots on the unit circle. For such polynomials one clearly has $M(f)=R$. 
Feb 6 
answered  Lower bound for spectral radius on $\operatorname{GL}(n,\mathbb{Z})$ 
Feb 3 
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Can there be arbitrarily many cubic fields unramified outside $\{p,\infty\}$?
"Silverman has proved that the last statement is implied by the ABC conjecture joint with the boundedness of MordellWeil ranks." This is actually a joint result with Marc Hindry. 
Jan 28 
answered  Proof of a Proposition regarding the reduction of Ntorsion groups on elliptic curves 
Jan 26 
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A morphism of elliptic schemes that preserves the identity is a homomorphism
Although it's just over fields, the proof that this is true for abelian varieties given in Mumford's Abelian Varieties via a rigidity argument is also quite enlightening. 
Jan 23 
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Elliptic curves and connected components
@StevenStadnicki Not sure if this will satisfy you, but $E(\mathbb{R})$ is a compact real Lie group, so it's realanalytically isomorphic to $S^1\times\Phi$, where~$S^1$ is the circle group and $\Phi$ is a finite group. (For elliptic curves, of course, $\Phi$ has to have order 1 or 2.) Anyway, the connected component of $E(\mathbb{R})$ corresponds to the group $S^1$ in this identification. 
Jan 21 
awarded  Enlightened 