Timeline for Diophantine equation: Egyptian fraction representations of 1
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Jun 30, 2010 at 8:31 | comment | added | Hugo van der Sanden | @Max: thanks, I've corrected it to -q (mod p). | |
| Jun 30, 2010 at 8:30 | history | edited | Hugo van der Sanden | CC BY-SA 2.5 |
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| Jun 30, 2010 at 0:02 | comment | added | Max Alekseyev | It should be $d\equiv -q\pmod{p}$, not $d\equiv -p\pmod{p}$. Basically, the underling trick here is that $$\frac{1}{u}+\frac{1}{u} = \frac{p}{q}$$ is reduced to a certain factorization of $q^2$: $$(pu-q)(pv-q) = q^2.$$ Assuming that the terms of expansion are sorted as $x_1 < x_2 < \dots < x_n$, they are typically generated from low-index to high-index and the above trick is used after the values to $x_1,\dots,x_{n-2}$ were assigned to quickly determine the values of $u=x_{n-1}$ and $v=x_n$ with $$\frac{p}{q} = 1 - \frac{1}{x_1} - \dots - \frac{1}{x_{n-2}}.$$ | |
| Jun 12, 2010 at 16:18 | vote | accept | Eric Rowell | ||
| Jun 12, 2010 at 16:18 | comment | added | Eric Rowell | Thanks! I really want the actual solutions and I was worried that the count was obtained in a significantly faster way. If I understand one would have to follow Eppstein's approach to get down to determining $x_1$ and $x_2$ and then one can count. I guess I should try to get some time on my university's supercomputer! | |
| Jun 12, 2010 at 10:29 | history | answered | Hugo van der Sanden | CC BY-SA 2.5 |