V p ( a 2 + 1 , a 2 ) ≡ V 1 ( a 2 + 1 , a 2 ) {\displaystyle V_{p}(a_{2}+1,a_{2})\equiv V_{1}(a_{2}+1,a_{2})} is equal to a 2 p ≡ a 2 mod p {\displaystyle a_{2}^{p}\equiv a_{2}\mod p}
Proof:
a = P + P 2 − 4 Q 2 {\displaystyle a={\frac {P+{\sqrt {P^{2}-4Q}}}{2}}} and b = P − P 2 − 4 Q 2 {\displaystyle b={\frac {P-{\sqrt {P^{2}-4Q}}}{2}}}
P = ( a 2 + 1 ) {\displaystyle P=(a_{2}+1)\ } and Q = a 2 {\displaystyle Q=a_{2}\ }
a = ( a 2 + 1 ) + ( a 2 + 1 ) 2 − 4 a 2 2 = 2 a 2 2 = a 2 {\displaystyle a={\frac {(a_{2}+1)+{\sqrt {(a_{2}+1)^{2}-4a_{2}}}}{2}}={\frac {2a_{2}}{2}}=a_{2}}
b = ( a 2 + 1 ) − ( a 2 + 1 ) 2 − 4 a 2 2 = 2 2 = 1 {\displaystyle b={\frac {(a_{2}+1)-{\sqrt {(a_{2}+1)^{2}-4a_{2}}}}{2}}={\frac {2}{2}}=1}
Follows:
V n ( a 2 + 1 , a ) = a 2 n + 1 {\displaystyle V_{n}(a_{2}+1,a)=a_{2}^{n}+1\ }
follows:
V p ( a 2 + 1 , a ) − V 1 ( a 2 + 1 , a ) = a 2 p + 1 − ( a 2 + 1 ) = a 2 p − a 2 {\displaystyle V_{p}(a_{2}+1,a)-V_{1}(a_{2}+1,a)=a_{2}^{p}+1-(a_{2}+1)=a_{2}^{p}-a_{2}\ }
--arbol01 21:33, 15 November 2007 (CST)
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