Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Diophantine equations are polynomial equations $F=0$, or systems of polynomial equations $F_1=\ldots=F_k=0$, where $F,F_1,\ldots,F_k$ are polynomials in either $\mathbb{Z}[X_1,\ldots,X_n]$ of $\mathbb{Q}[X_1,\ldots,X_n]$ of which it is asked to find solutions over $\mathbb{Z}$ or $\mathbb{Q}$. Topics: Pell equations, quadratic forms, elliptic curves, abelian varieties, hyperelliptic curves, Thue equations, normic forms, K3 surfaces ...
4
votes
0
answers
453
views
Why is Hilbert’s 11th problem still partially resolved?
Hilbert’s 11th problem which demands that we ‘classify quadratic forms over algebraic number fields’ has been of interest to me and I would like to know what makes it partially resolved currently. Or …
-1
votes
Diophantine equation $2(x - 1/x) = y - 1/y$
When the Diophantine problem above is expressed as:
xy^2 + (2-2x^2)y - x = 0 (1)
and solved with respect to “y”, you would arrive at x=1 or x=-1 and consequently giving y=1 or y=-1 respectively when s …