March 19th, 2017, 11:32 AM  #1 
Newbie Joined: Mar 2017 From: sri lanka Posts: 3 Thanks: 0  if nx^2 then nx
1)if n is a positive integer and doesn't have a factor of any positive integer except 1 2)if x is an integer then, how to prove that if nx^2 then nx? 
March 19th, 2017, 01:17 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,876 Thanks: 2240 Math Focus: Mainly analysis and algebra 
1) You presumably mean "doesn't have a factor of any positive integer except $n$ and $1$". $n$ is prime, so you are going to refer to the prime factorisation of numbers, in particular that a prime factorisation is unique up to the order of the factors. 2) You might find it easier to prove the contrapositive: if $n$ does not divide $x$ then $n$ does not divide $x^2$. The two results are equivalent. 
March 19th, 2017, 01:29 PM  #3 
Newbie Joined: Mar 2017 From: sri lanka Posts: 3 Thanks: 0 
what if n can has factors of positive integers but not square factors?

March 19th, 2017, 04:12 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,876 Thanks: 2240 Math Focus: Mainly analysis and algebra 
$n$ is prime by the definition you gave.

April 5th, 2017, 03:57 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,576 Thanks: 668 
Your original statement is not true: 49 divides 49 but does not divide 7.

April 5th, 2017, 05:41 AM  #6  
Senior Member Joined: May 2016 From: USA Posts: 755 Thanks: 303  Quote:
I think the problem is $PROVE\ n\ is\ a\ prime\ and\ n\ \ x^2 \implies n\ \ x.$ I could be wrong about what is to be proved of course. EDIT: I just noticed the OP's second post. I now have no idea what he is asking. It appears now to deal with primes and composites that do not include powers, or at least powers of 2. Last edited by JeffM1; April 5th, 2017 at 05:46 AM.  
April 18th, 2017, 07:38 AM  #7 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,082 Thanks: 87 
pab > pa or pb Theorem in Birkhoff MacLane Last edited by zylo; April 18th, 2017 at 07:46 AM. 
April 18th, 2017, 09:33 AM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,876 Thanks: 2240 Math Focus: Mainly analysis and algebra  
April 18th, 2017, 01:36 PM  #9 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,347 Thanks: 224 
Since $n$ is prime it's prime factorization has a single factor raised to the first power. Denote $n$ as $p$ Now you can use the fact If $ \ \ p \ \  \ \ (a \times b) \ \ $ Then $ \ \ p \ \  \ \ a \ \ $ Or $ \ \ p \ \  \ \ b $ $x^2 = x \times x$ Now it's easy , $p \ \  \ \ (x \times x)$ Which means $p \ \  \ \ x \ \ $ Or $ \ \ p \ \  \ \ x$ You win either way 
April 18th, 2017, 07:23 PM  #10  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,347 Thanks: 224  Quote:
Let $ \ \ n = p_1 \times p_2 \times ... \times p_k $ Given $( p_1 \times p_2 \times ... \times p_k ) \ \  \ \ (x \times x )$ If any $ \ \ p_i \ \ $ divides any $ \ \ x \ \ $ then it must divide the other $ \ \ x \ \ $ as well. (Doubly confirmed ) You win again ...  