Timeline for fibonacci series mod a number
Current License: CC BY-SA 2.5
4 events
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| Nov 7, 2010 at 19:52 | comment | added | Nick S | Also note that the order of $GL_2(\Z_k)$ is much smaller than $n$, and the order of the matrix divides this order. If $l$ is the reminder of $n$ divided by this order, then $A^n=A^l \mod p$. Last but not least it is enough to consider the subgroup of matrices of $det =\pm1$. | |
| Oct 4, 2010 at 12:28 | comment | added | user9734 | Sorry for the late accept, I have been busy lately. Thank you all for the help! | |
| Oct 4, 2010 at 12:28 | vote | accept | user9734 | ||
| Oct 2, 2010 at 6:51 | history | answered | Robin Chapman | CC BY-SA 2.5 |