My Math Forum if n|x^2 then n|x

 Real Analysis Real Analysis Math Forum

 March 19th, 2017, 11:32 AM #1 Newbie   Joined: Mar 2017 From: sri lanka Posts: 3 Thanks: 0 if n|x^2 then n|x 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 n|x^2 then n|x?
 March 19th, 2017, 01:17 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,671 Thanks: 2171 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 contra-positive: if $n$ does not divide $x$ then $n$ does not divide $x^2$. The two results are equivalent. Thanks from topsquark and Singi
 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: 6,671 Thanks: 2171 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: 2,403 Thanks: 608 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: 591
Thanks: 251

Quote:
 Originally Posted by Country Boy Your original statement is not true: 49 divides 49 but does not divide 7.
I think we have to presume that the OP meant to limit n to primes. 49 is not a prime.

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: 1,006 Thanks: 81 p|ab -> p|a or p|b Theorem in Birkhoff MacLane Thanks from agentredlum and topsquark 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: 6,671
Thanks: 2171

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by Country Boy Your original statement is not true: 49 divides 49 but does not divide 7.
But 7 divides 49, so $n \ne 49$.

 April 18th, 2017, 01:36 PM #9 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,217 Thanks: 185 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 Thanks from v8archie
April 18th, 2017, 07:23 PM   #10
Math Team

Joined: Jul 2011
From: North America, 42nd parallel

Posts: 3,217
Thanks: 185

Quote:
 Originally Posted by Singi what if n can has factors of positive integers but not square factors?
I think you are saying that the prime factorization of $n$ has all of it's prime factors to the first power.

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 ...

 Tags n|x, n|x2

 Thread Tools Display Modes Linear Mode

 Contact - Home - Forums - Cryptocurrency Forum - Top