for $\displaystyle n \in \mathbb{N} $ Let $\displaystyle \ p_n $ \ be the $\displaystyle n^{th}$ prime $\displaystyle (thus \ , p_1=2,p_2=3,p_3=5,p_4=7,p_5=11,...):$

prove that $\displaystyle \ p_n \leq p_1 p_2 p_3...p_{n-1} +1 , n \geq 3$

Printable View

- Mar 1st 2010, 02:44 AMflower3prove,,,
for $\displaystyle n \in \mathbb{N} $ Let $\displaystyle \ p_n $ \ be the $\displaystyle n^{th}$ prime $\displaystyle (thus \ , p_1=2,p_2=3,p_3=5,p_4=7,p_5=11,...):$

prove that $\displaystyle \ p_n \leq p_1 p_2 p_3...p_{n-1} +1 , n \geq 3$ - Mar 1st 2010, 09:02 AMqmech
Suppose the next prime wasn't less or equal to your number (A= product of primes + 1). Is A prime? Clearly none of the primes less than A divide it (they all leave a remainder of 1). You have a contradiction.