# Proof with prime numbers

• Oct 25th 2009, 07:46 PM
MichaelG
Proof with prime numbers
Given that p doesn't divide n for all primes p is less than equal to the cube root of n, show that n> is either a prime or the product of two primes.

I was given a hint and I was told to assume to the contrary that n contains at least 3 prime factors.
• Oct 25th 2009, 09:58 PM
NonCommAlg
Quote:

Originally Posted by MichaelG
Given that p doesn't divide n for all primes p is less than equal to the cube root of n, show that n> is either a prime or the product of two primes.

I was given a hint and I was told to assume to the contrary that n contains at least 3 prime factors.

suppose $n=\prod_{i=1}^k p_i,$ where $k \geq 3$ and $p_1 \leq p_2 \leq \cdots \leq p_k$ are primes. then $p_1^3 \leq p_1^k \leq \prod_{i=1}^k p_i = n.$ contradiction!