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January 19th, 2018, 10:01 PM  #1 
Member Joined: Sep 2011 Posts: 99 Thanks: 1  Modular Arithmetic proof
Hi, I have encountered some difficulties with the question in the zip file attached. Greatly appreciate any help given! Thanks in advance.

January 19th, 2018, 11:09 PM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,764 Thanks: 621 Math Focus: Yet to find out. 
Can you upload an image instead?

January 19th, 2018, 11:15 PM  #3 
Member Joined: Sep 2011 Posts: 99 Thanks: 1 
I have uploaded the image instead. thanks.

January 20th, 2018, 02:27 AM  #4 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 300 Thanks: 96 Math Focus: Number Theory, Algebraic Geometry  This is just using the language of modular arithmetic to state a (hopefully) familiar result. It's just saying: 1) Show that if $p$ is a prime and $p$ divides $ab$ then $p$ divides $a$ or $p$ divides $b$. 2) Give a counterexample to the above statement if $p$ is composite. 
January 20th, 2018, 07:56 AM  #5 
Senior Member Joined: Sep 2016 From: USA Posts: 555 Thanks: 319 Math Focus: Dynamical systems, analytic function theory, numerics 
To expand a bit on #1 from cjem's reply: Suppose $A \in [a], B \in [b]$, and write $A = mp+a,B = np+b$ for some $m,n \in \mathbb{Z}$. Then, the statement that $[a][b] = [0]$ simply means that $p$ divides $AB = nmp + npa + mpb + ab$. The first 3 terms are clearly multiples of $p$ so then the last term is required to be a multiple of $p$ as well which leads to the requirement that $p$ divides $ab$. The conclusion that $[a]$ or $[b]$ equals $[0]$ means that either $a$ or $b$ is a multiple of $p$. Taken together, the claim is that if $p$ divides $ab$, then $p$ divides either $a$ or $b$ as was pointed out by cjem. 

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