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Nov
25 |
comment |
Deep/precise relationship between two approaches to FLT for polynomials, $n = 3$
It can be helpful to see the difference in the arithmetic side. The first proof reminds me Kummer's proof of FLT when the exponent is a regular prime (or even better, when unique factorization holds). The second proof corresponds to deducing FLT from ABC in the arithmetic side (R-H formula for function fields and the truncated second main theorem for meromorphic functions are usually seen as analogues of ABC). One approach works in many cases over Z, the other remains incomplete as far as I know. |
Sep
5 |
awarded | Yearling |
Aug
24 |
comment |
Philosophy behind Mochizuki's work on the ABC conjecture
Apparently, this answer written in September 2012 turned out to be not-so-unrelated to the question. See kurims.kyoto-u.ac.jp/~motizuki/… |
Aug
17 |
comment |
For which rings $A$ is $A-\{0\}$ Diophantine over $A$?
Read the following paper by L. Moret-Bailly: Sur la définissabilité existentielle de la non-nullité dans les anneaux, Algebra and Number Theory. vol. 1, n° 3 (2007), 331-346. |
Jul
29 |
awarded | Autobiographer |
Jul
29 |
revised |
Genus of a plane curve of the form $\prod_{i=1}^n (a_iX+b_iY+Z) = Z^n$
edited body |
Jul
29 |
answered | Genus of a plane curve of the form $\prod_{i=1}^n (a_iX+b_iY+Z) = Z^n$ |
Jul
22 |
comment |
When is the image of an integral polynomial contained in the image of another?
I should mention that the idea in this argument is a particular case of the previously mentioned paper of Davenport-Lewis-Schinzel, where a more general result is proved using HIT. |
Jul
21 |
comment |
When is the image of an integral polynomial contained in the image of another?
I think that using the suggested results will only lead to complicated issues of rationality and a lengthy case-by-case analysis on the pairs in the paper of Bilu and Tichy. See Theorem 1 in: H. Davenport, D. J. Lewis, A. Schinzel, Polynomials of certain special types. Acta Arith. 9 (1964), 107-116 for a result better suited for this problem. |
Apr
20 |
awarded | Enthusiast |
Apr
15 |
comment |
Generalization of proposition of Granville related to abc conjecture
What my comment hints is a way to prove the bound with $+1$ (using what one knows in the 1-variable case). You understood something different, but this misundertanding is actually interesting (unless I misunderstood my own comment). |
Apr
13 |
comment |
Generalization of proposition of Granville related to abc conjecture
Except for the "+2" (which requires finer analysis on pencils in $\mathbb{P}^2$) a bound of this sort might be proved as follows: a general linear substitution $u=at+b$, $v=ct+d$ will preserve the degree of $r,s$, the coprimality of $r,s$, and cannot increase the degree of $rad(G(r,s))$. |
Apr
4 |
answered | Decidability of sum of powers exponential diophantine equation |
Feb
7 |
comment |
Paths in $\mathrm{Spec} \, \mathbb{Z}$ and Kim's proof of Siegel's theorem for $\mathbb{P}^1 \setminus \{0,1,\infty\}$
As of today, I don't think that such a bound is known to follow from Kim's proof. Nevertheless, there is some work on computing the various objects involved in the proof, in order to get effective finiteness, see for instance 1209.0640v3 |
Feb
3 |
comment |
Paths in $\mathrm{Spec} \, \mathbb{Z}$ and Kim's proof of Siegel's theorem for $\mathbb{P}^1 \setminus \{0,1,\infty\}$
About clarifying the idea of paths: did you try reading this? people.maths.ox.ac.uk/kimm/papers/leeds.pdf |
Nov
19 |
comment |
Is there a general connection between value distribution and zero distribution for functions representable by Dirichlet series?
Possibly unrelated comment: If instead of strips you use discs of increasing radii then the property that you want is false but it has a reasonable fix: add proximity functions and use the first main theorem of Nevanlinna theory. If you allow different (fixed, finitely many, at least 3) choices of z then you don't need proximity functions (second main theorem). |
Nov
11 |
awarded | Nice Answer |
Nov
10 |
answered | Images of polynomials |
Nov
10 |
comment |
Pairs of quadratic polynomials taking values pairs of consecutive squares
Interesting. Have you tried to do something? Do you have any guess? |
Nov
6 |
answered | Faltings height in short exact sequences |