Talk:Lucas sequence

Equality

$V_{p}(a_{2}+1,a_{2})\equiv V_{1}(a_{2}+1,a_{2}){\pmod {p}}$ is equal to $a_{2}^{p}\equiv a_{2}{\pmod {p}}$

Proof:

$a={\frac {P+{\sqrt {P^{2}-4Q}}}{2}}$ and $b={\frac {P-{\sqrt {P^{2}-4Q}}}{2}}$

$P=(a_{2}+1)\$ and $Q=a_{2}\$

$a={\frac {(a_{2}+1)+{\sqrt {(a_{2}+1)^{2}-4a_{2}}}}{2}}={\frac {2a_{2}}{2}}=a_{2}$

$b={\frac {(a_{2}+1)-{\sqrt {(a_{2}+1)^{2}-4a_{2}}}}{2}}={\frac {2}{2}}=1$

Follows:

$V_{n}(a_{2}+1,a_{2})=a_{2}^{n}+1\$

follows:

$V_{p}(a_{2}+1,a_{2})-V_{1}(a_{2}+1,a_{2})=a_{2}^{p}+1-(a_{2}+1)=a_{2}^{p}-a_{2}\$

--arbol01 21:33, 15 November 2007 (CST)

Improvement of readability

Please do not be offended, but this article was not so readable due to the poor use of English. I've tried to help improve the presentation of this article. --Hendra 07:34, 17 November 2007 (CST)

Oh, i am not offended. I am german, and i know about my foibles. I am very grateful to every help, i can get. --arbol01 08:03, 17 November 2007 (CST)