October 6th, 2016, 11:30 PM  #1 
Member Joined: Sep 2016 From: zambia Posts: 31 Thanks: 0  irrational numbers
1. Show that if p^2 is divisible by 3,then p is also divisible by 3

October 7th, 2016, 06:23 AM  #2 
Member Joined: Aug 2015 From: Montenegro (Podgorica) Posts: 37 Thanks: 4 
for p^2 to be devisible by 3, p needs to be 3 or composite number that have 3 at least once in his prime factors..

October 7th, 2016, 06:40 AM  #3 
Senior Member Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 
If p is divisible by 3, then: p=3n Let's make prime factorization of p: $\displaystyle p=a_1a_2a_3...a_m$ Then: $\displaystyle p^2=a_1a_1a_2a_2a_3a_3...a_ma_m$ Let's make prime factorization of 3n: $\displaystyle 3b_1b_2b_3...b_{2m1}$ 2m1 is because of unique factorization theorem, and because of the same theorem, one of the $\displaystyle a_1,a_2,a_3,...,a_m$ must equal to 3. So: $\displaystyle p=a_1a_2a_3...\cdot 3\cdot...a_m$ is divisible by 3. 
October 7th, 2016, 07:05 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra  
October 8th, 2016, 01:12 AM  #5 
Member Joined: Sep 2016 From: India Posts: 88 Thanks: 30 
If $p$ is divisible by 3 then it will also divisible by $p^2$ and $p^2$ is divisible by 3 then it will also divisible by $p$.

October 8th, 2016, 05:59 AM  #6  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
Because of the ambiguous construction, the pronoun, "it," which I highlighted, could refer to p or to 3. If "it" refers to p, then part of your statement means "p will also [be] divisible by $\displaystyle \ p^2. \ \ $ That isn't true, unless p = $\displaystyle \ p^2 \ \ne \ 0. $ If "it" refers to 3, then that part of your statement means "3 will also [be] divisible by $\displaystyle \ p^2. \ \ $ That isn't true unless p is 1 or 1. Last edited by Math Message Board tutor; October 8th, 2016 at 06:15 AM.  

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