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March 19th, 2017, 11:32 AM   #1
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Unhappy 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?
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March 19th, 2017, 01:17 PM   #2
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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.
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March 19th, 2017, 01:29 PM   #3
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what if n can has factors of positive integers but not square factors?
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March 19th, 2017, 04:12 PM   #4
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$n$ is prime by the definition you gave.
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April 5th, 2017, 03:57 AM   #5
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Your original statement is not true: 49 divides 49 but does not divide 7.
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April 5th, 2017, 05:41 AM   #6
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Quote:
Originally Posted by Country Boy View Post
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.
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April 18th, 2017, 07:38 AM   #7
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p|ab -> p|a or p|b

Theorem in Birkhoff MacLane
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Last edited by zylo; April 18th, 2017 at 07:46 AM.
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April 18th, 2017, 09:33 AM   #8
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Quote:
Originally Posted by Country Boy View Post
Your original statement is not true: 49 divides 49 but does not divide 7.
But 7 divides 49, so $n \ne 49$.
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April 18th, 2017, 01:36 PM   #9
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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

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April 18th, 2017, 07:23 PM   #10
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Quote:
Originally Posted by Singi View Post
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 ...

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