bio  website  mathematik.uniwuerzburg.de/… 

location  Würzburg  
age  49  
visits  member for  3 years, 9 months 
seen  2 hours ago  
stats  profile views  2,281 
I'm a professor of mathematics at the University of Würzburg (Germany). For further infos see my web page.
20h

comment 
Weierstrass form of genus one $y^{10} z^{30}  8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10}  64=0$
@joro: What do you mean? In your question, you ask for a Weierstrass form! So what does your comment mean that you have a Weierstrass form which is birationally equivalent to the given curve? By the way, I still believe that I correctly answered your question. 
21h

revised 
Weierstrass form of genus one $y^{10} z^{30}  8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10}  64=0$
added 482 characters in body 
22h

comment 
Weierstrass form of genus one $y^{10} z^{30}  8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10}  64=0$
A nonsingular curve given in Weierstrass form is absolutely irreducible, hence so is every curve which is birationally equivalent (no matter over which field) to this curve. But your curve factors over $\mathbb Q(\sqrt[5]{2})$. 
22h

comment 
Weierstrass form of genus one $y^{10} z^{30}  8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10}  64=0$
@joro: I just noticed that your curve isn't absolutely irreducible, while it is reducible over the rationals! So that tremendously simplifies things, because only singularities can be rational points! (But non of them, except for two points at infinity, are rational.) I guess this also explains the error messages in Maple when trying to find the Weierstrassform, because the program certainly expects an absolutely irreducible curve of genus $1$. 
23h

comment 
Weierstrass form of genus one $y^{10} z^{30}  8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10}  64=0$
@joro: You are right, $C$ indeed has only finitely many rational points. I amended my answer. 
23h

revised 
Weierstrass form of genus one $y^{10} z^{30}  8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10}  64=0$
added 816 characters in body 
1d

comment 
Genus of a plane curve of the form $\prod_{i=1}^n (a_iX+b_iY+Z) = Z^n$
@Jason Starr: In which sense can, for $n\ge3$, the $n$ pairs $(\bar a_i,\bar b_i)\in\overline{\mathbb F_q}^2$ be linearly independent? 
1d

answered  Genus of a plane curve of the form $\prod_{i=1}^n (a_iX+b_iY+Z) = Z^n$ 
1d

revised 
Weierstrass form of genus one $y^{10} z^{30}  8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10}  64=0$
added 170 characters in body 
2d

revised 
Weierstrass form of genus one $y^{10} z^{30}  8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10}  64=0$
added 262 characters in body 
2d

answered  Weierstrass form of genus one $y^{10} z^{30}  8000 y^{4} z^{20} + 12800000 z^{20} + 1600 y^{2} z^{10}  64=0$ 
Jul 22 
awarded  Enlightened 
Jul 22 
comment 
When is the image of an integral polynomial contained in the image of another?
@David: I tried to clarify my argument. The point is that for some fixed $i$, $A_i(h,Y)$ is irreducible and at the same time has a rational root for infinitely many integers $h$. And this of course implies $Y$degree $1$. 
Jul 22 
revised 
When is the image of an integral polynomial contained in the image of another?
added 85 characters in body 
Jul 22 
awarded  Nice Answer 
Jul 22 
answered  When is the image of an integral polynomial contained in the image of another? 
Jul 15 
comment 
Upper density of the set of $n$'s such that $p(n)$ is prime, where $p$ is polynomial
Does anyone of those who downvoted the question or opted to close it care to tell the reason? To me it seems a reasonable question, and despite some of the comments it seems that an accurate answer isn't that easy. 
Jun 30 
comment 
Need explicit formula for certain “$q$numbers” involving gcd's
@OP: What is $\langle e_0,e_0\rangle$? It seems to me that the problem is not well formulated, as $gcd(0,0)=\infty$. In particular, I don't know how to interpret your assertion $0=\langle o_0,o_1\rangle=\langle e_0,e_1qe_0\rangle=qq\cdot q^{gcd(0,0)}$. 
Jun 21 
comment 
How to prove this polynomial always has integer values at all integers?
Comparing the leading coefficient of $P_m(x)$ with that of $\binom{x}{2m}$, one gets that $\frac{3}{(2m+1)(m1)}\binom{2m}{m}\sum_{i=0}^m\binom{m}{i}^2\frac{1}{2i1}$ is an integer provided that $P_m(\mathbb Z)\subseteq\mathbb Z$. Is this integrality known? I don't know if there is a closed expression for the sum. 
Jun 8 
comment 
Existence of polynomials of degree $\geq 2$ which represent infinitely many prime numbers
@Gerhard Paseman: $x^2+y^2$ even assumes all the infinitely many primes in $4\mathbb N+1$ ... 