Timeline for Diophantine equation problem
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 2, 2010 at 12:18 | comment | added | Wadim Zudilin | Cindy, why don't you guess the fourth nonnegative solution $(0,0,1)$? (In my comment to Will's post below I explain how to give all rational points on your ellipsoid. It is this fourth solution through which the lines with rational pair of slopes pass.) | |
| S May 31, 2010 at 1:17 | vote | accept | MathematicianMummy | ||
| S May 31, 2010 at 1:17 | vote | accept | MathematicianMummy | ||
| S May 31, 2010 at 1:17 | |||||
| May 31, 2010 at 1:13 | vote | accept | MathematicianMummy | ||
| S May 31, 2010 at 1:17 | |||||
| May 30, 2010 at 19:18 | answer | added | Will Jagy | timeline score: 1 | |
| May 30, 2010 at 18:45 | answer | added | Kevin O'Bryant | timeline score: 3 | |
| May 30, 2010 at 18:30 | comment | added | Pietro Majer | Just to clarify before closing: write your equation (x-z/2)^2 + (y-z/2)^2 + z^2/2 =1 , whence z is either 0 or 1, &c. | |
| May 30, 2010 at 18:20 | comment | added | S. Carnahan♦ | You may want to change "positive" to "non-negative" in your question, and you may want to consider solutions with more zeroes. I'm afraid I'm voting to close, since your question is a little outside the scope of this site. Please see the FAQ for a list of problem-solving websites. | |
| May 30, 2010 at 17:27 | comment | added | Robin Chapman | That's the equation of an ellipsoid: a bounded set. Diagonalizing it will give explicit bounds for $x$, $y$ and $z$. By diagonalization I mean a change of coordinates transforming it to the form $ax^2+by^2+cz^2=d$. | |
| May 30, 2010 at 17:20 | history | asked | MathematicianMummy | CC BY-SA 2.5 |