# Noam D. Elkies

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## Registered User

 Name Noam D. Elkies Member for 2 years Seen 6 hours ago Website Location Harvard University, Cambridge, MA 02138 Age 46
[Not sure yet what to use this space for...]
 2d comment Real root of a cubic equationThe canonical answer is Sturm sequences (see for example the Wikipedia entry en.wikipedia.org/wiki/Sturm%27s_theorem ), which let one count the roots of a polynomial of any degree in ${\bf R}$ or in an interval. I might have written more about it, but the question was already closed so this will have to do. May21 accepted Elliptic curves over QQ with identical 13-isogeny May19 comment Good uses of Siegel zeros?Is this really a formula for $h^-(p)$, not $\log h^-(p)$? May19 comment Lipschitz map of the ellipseThe OP must want it to be Lipschitz with respect to the metric induced from the Euclidean plane, not the intrinsic Riemannian metric on the ellipse and circle. It still feels like the answer should still be yes but it's not as immediate. May19 comment Why are affine Lie algebras called affine?I always assumed it was because they correspond to affine reflection groups in the way that finite-dimensional semisimple algebras correspond to spherical (Euclidean) reflection groups. May18 answered Sequences equidistributed modulo 1 May17 comment maximizing multivariate polynomialThis seems more suitable for artofproblemsolving, but anyway: Since the polynomial is linear in each variable $y_i$ separately (once $N>2$), we may assume each $y_i \in \lbrace 0, 1 \rbrace$ (even if the intention was to just limit to the hypercube $0 \leq y_i \leq 1$. So we're just asking for the maximal number of 010 patterns in a cycle of $N+1$ zeros and ones. Since no two consecutive triples can be 010, the count is at most $\lfloor (N+1)/2 \rfloor$, and this is easily attained, and only by 010101... and its cyclic shifts (a total of $2$ if $N$ is even and $N+1$ if $N$ is odd). May17 comment Fano plane drawings: embedding PG(2,2) into the real planeActually I'm not using anything lke that (certainly not for an arbitrary field). Another way to say this is to choose projective coordinates so $A$, $B$, and $C$ are at the unit vectors $(1:0:0)$, $(0:1:0)$, and $(0:0:1)$, and then scale those coordinates so that $O$, which must have all three coordinates nonzero (else it's on one of the lines $AB$, $AC$, $BC$) is on $(1:1:1)$; then $OA$ is the line $y=z$, so $A'=OA \cap BC$ is $(0:1:1)$, and likewise $B = (1:0:1)$ and $C = (0:1:1)$. Now calculate that the determinant of $A,B,C$ is $2$, so $ABC$ are collinear iff we're in characteristic 2. May17 revised Elliptic curves over QQ with identical 13-isogenyInclude (t,X) = (3,-115/126) curves May16 answered Elliptic curves over QQ with identical 13-isogeny May15 comment Fano plane drawings: embedding PG(2,2) into the real plane...and conversely, if $k$ does have characteristic $2$ then $A',B',C'$ are always collinear... May15 comment Fano plane drawings: embedding PG(2,2) into the real planeYes, a line drawing is impossible, over ${\bf R}$ or any field $k$ not of characteristic $2$. Let $A,B,C,O$ be non-collinear points of the Fano plane, and $A',B',C'$ the intersections of $AO,BO,CO$ with $BC,CA,AB$ respectively. By Ceva's theorem (actually proved by Al-Mutaman centuries earlier, and extended algebraically to the case where $O$ is outside the triangle, and indeed to arbitrary $k$), points $A',B',C'$ divide segments $BC,CA,AB$ in signed ratios whose product is $1$. But by Menealus' theorem, $A',B',C'$ are collinear iff that product is $-1$. Since $1 \neq -1$ we're done. May10 comment Closed form for derivatives $\zeta^{(n)}(1/2)$Actually you only get every other derivative for free this way. The functional equation says $\xi(s) = \pi^{s/2} \Gamma(s/2) \zeta(s)$ is symmetric about $s = 1/2$, so its odd-order derivatives vanish there, which gives linear equations on the $\zeta^{(n)}(1/2)$ that let you solve for $\zeta^{(2m+1)}(1/2)$ as a linear combination of $\zeta(1/2)$, $\zeta''(1/2)$, $\zeta^{(4)}(1/2)$, ..., $\zeta^{(2m)}(1/2)$. But you still can't solve for the even-order derivatives in terms of derivatives of lower order. May9 awarded ● Nice Answer May9 comment Prove that the sum of a certain infinite series is 1Then too, some numerically evident propositions are false... For example, $$\sum_{n=1}^\infty \frac{(3n-2)!}{(2n)!} \frac{(2n+99)!}{(3n+99)!}$$ is not quite equal to what numerical computation suggests. See math.harvard.edu/~elkies/Misc/sol11.html . May9 comment What is known about a^2 + b^2 = c^2 + d^2But the last puzzle was already solved by Fermat...  ...Which also suggests an alternative approach to the equation $a^2+b^2=c^2+d^2$ starting from the factorization $(a+bi)(a-bi)=(c+di)(c-di)$ in ${\bf Z}[i]$. May9 comment What is known about a^2 + b^2 = c^2 + d^2$a^2+b^2=c^2+d^2 \Longleftrightarrow a^2-c^2=d^2-b^2 \Longleftrightarrow (a-c)(a+c)=(d-b)(d+b)$. Now the primitive solutions of $rs=tu$ are exactly $(r,s,t,u) = (xx',yy',xy',x'y)$ with $\gcd(x,y)=\gcd(x',y')=1$ (and some positivity condition to avoid duplication with factors of $-1$). If $(a,b,c,d)$ is primitive then $(r,s,t,u)$ is primitive up to a factor of $2$ and satisfies $r\equiv s$ and $t\equiv u \bmod 2$. So just figure out what to do with $x,x',y,y' \bmod 2$ and you're done. May7 revised cube + cube + cube = cubeSupply apostrophe in "can't" May7 comment cube + cube + cube = cubeYes, JHI, as I wrote correctly in my actual answer; here I must have been distracted by the alphabetical progression $-$ sorry. Unlike answers, comments cannot be edited here except by deleting the original comment and entering the corrected text for a new comment... May6 comment cube + cube + cube = cubeThanks too for accept ing my new answer. This curiously gives GHI the rare, perhaps unique, distinction of earning a gold Populist badge for his first and only MO answer! Populist or not, GHI's answer is necessary to show that the 8-piece dissection is minimal, and greatly helped focus my search using the key condition that each piece much include a corner of the $6 \times 6 \times 6$ cube. It also has the virtue of applying to all $(a,b,c;d)$, not just $(3,4,5;6)$.  While I'm at it: who is "GHI"? When citing his contribution I'd like to credit a real person rather than a pseudonym. May6 awarded ● Good Answer May6 revised cube + cube + cube = cubeFix/improve some sentences; give lower bound of 9 on taxicab dissection May6 comment cube + cube + cube = cubeThanks! The 2-D illustration was possible because it's how I was able to find the dissection in the first place $-$ fortunately no coordinate permutation was needed except for boxes. It would have been much harder both to find and to illustrate the dissection if some of the pieces were as complicated as in the 9-piece dissection that you mentioned and F.Brunault posted. May6 accepted cube + cube + cube = cube May5 awarded ● Nice Answer May5 awarded ● Necromancer May5 comment Prove a+b+c+d is composite.Geometrically $ab=cd$ is a split quadric in ${\bf P}^3$, so isomorphic with ${\bf P}^1 \times {\bf P^1}$, and then the $(1,1)$ form $a+b+c+d$ factors as a product of a $(1,0)$ and a $(0,1)$, etc. But I doubt that this is the kind of solution "CODE" is looking for... May5 comment Prove a+b+c+d is composite.Off topic here, but good for artofproblemsolving. May5 answered cube + cube + cube = cube May3 awarded ● Enlightened May3 awarded ● Nice Answer May2 comment Why is it hard to prove that the Euler Mascheroni constant is irrational?Note, though, that the fact that $\gamma$ is not known to be a "period" does not exclude an irrationality proof from some other direction; the irrationality of numbers such as $\log_2 3$ is even easier to prove than the irrationality of $\pi$, and $\log_2(3)$ is not expected to be a period (though it's the ratio of the periods $\log 3$ and $\log 2$). May1 awarded ● Yearling Apr29 comment Periods and commas in mathematical writing"Tilde" does come from "Medieval Latin titulus tittle" according to m-w.com; but this "jot" is not the usual "to write briefly or hurriedly" but a variant of "iota". So, "every $\iota$ and tilde"? Apr25 answered The integral inequality Apr23 comment Resources on Wolstenholme’s theorem...or Google: the first Google hit for glaisher wolstenholme is ;, where on page 25 there are three references ([32] to [34]) to papers by Glaisher published in the Q.J.Math. in 1900 or 1901; the second of these ("On the residues of the sums of products of the first $p - 1$ numbers, and their powers, to modulus $p^2$ or $p^3$") seems particularly promising. Apr22 comment Solutions to $\binom{n}{5} = 2 \binom{m}{5}$Using Stahlke and Stoll's program ratpoints, I find that there are no further rational solutions of $u^2 = 9t^6+16t^5-200t^3+256t+144$ with $u=r/s$ and $|r|,|s| \leq 10^5$. On the original equation, it should be easy to show there are no further solutions up to say $10^{100}$ using the fact that $(n-2)/(m-2) = 2^{1/5} + O(1/m)$, so there are only a few hundred approximations to $2^{1/5}$ that are close enough, and they can be computed via the continued-fraction expansion. Apr12 awarded ● Enlightened Apr12 awarded ● Nice Answer Apr12 comment Numbers integrally represented by a ternary cubic formYou're welcome, and thank you for accepting my answer. Apr12 revised Numbers integrally represented by a ternary cubic formAdd paragraph on Minkowski bound Apr11 accepted Numbers integrally represented by a ternary cubic form Apr11 answered Numbers integrally represented by a ternary cubic form Apr11 comment Numbers integrally represented by a ternary cubic formSure, there are large powers with primitive representations. This follows from the multiplicativity that Will J. already noted. For instance, $f(3356898, 3732782, 5764967) = 7^{23}$ (I started from $f(2,1,0)=7$ and used $7^{23}=1^{10}7^{23}$ and $f(2,-1,0)=1$ to reduce the resulting $(a,b,c)=(472709258936428, 396738620092614, 257006830281609)$ ). Apr10 comment Lower bounds for Petersson inner products of cuspforms with integral Fourier coefficientsAh, I saw the "integer coefficients" part but didn't appreciate the significance of "newform" (implying not just in the "new" space but an actual Hecke eigenform). Apr9 comment Lower bounds for Petersson inner products of cuspforms with integral Fourier coefficientsGiven $f$, the possible $(f,g)$ form a subgroup of ${\bf R}$, which is either discrete or dense. Once the space of cuspforms has dimension at least $2$ one would expect it to be dense unless $f = 0$ (why should two or more "random" Petersson products be proportional?), and thus to contain arbitrarily small positive elements. Is there a further missing assumption? Apr8 comment $a^5+b^5=c^5+d^5$ and polynomial identitiesThat genus-zero parametrization gets rediscovered often. It's actually defined over ${\bf Q}$ but alas has no rational points, being birational with the conic $x^2+y^2+z^2=0$. The nice way to see it is to intersect the quintic surface $a^5+b^5+c^5+d^5=0$ with the line $a+b+c+d=0$: the resulting quintic curve decomposes into three lines $a+b=c+d=0$, $a+c=d+b=0$, $a+d=b+c=0$, and a residual conic with $S_4$ symmetry that has no rational or even real roots but does work over ${\bf Q}(i)$ and various other totally imaginary number fields. Apr8 answered Characterise all pairs of n/m stars that have the same inner radius Apr7 awarded ● Nice Answer Apr7 revised sum of three cubes and parametric solutionsfix two minor typos