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Inegalitate cu functia prime-counter pi
Posted: Tue Oct 09, 2007 12:50 am
by Cezar Lupu
Sa se arate ca daca \( \pi(x) \) reprezinta numarul numerelor prime mai mici ca un numar \( x \) atunci are loc inegalitatea
\( \pi(n!+2n)+\pi(n)\leq\pi(n!+n)+\pi(2n), \forall n\geq 1 \).
Posted: Tue Oct 09, 2007 10:15 am
by Dragos Fratila
It is equivalent to prove that a prime between \( n!+n \) and \( n!+2n \) remains prime when we take it modulo \( n! \).
Suppose it doesn't. Then it must have a divisor less than or equal to \( n \). That is \( p=n!+q\cdot k \), where \( q \) is a prime less than \( n \). Then \( p =0 \) modulo \( q \) since \( q \) divides \( n! \), hence \( p \) is not prime. Contradiction.
ma scuzati va rog ca am scris in engleza ... citeam ceva in engleza cand mi-a venit ideea... abia dupa ce am terminat m-am prins, si nu mai rescriu acu... scuze inca o data