My Math Forum Modular Arithmetic proof

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January 19th, 2018, 10:01 PM   #1
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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.
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 January 19th, 2018, 11:09 PM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,734 Thanks: 605 Math Focus: Yet to find out. Can you upload an image instead?
January 19th, 2018, 11:15 PM   #3
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January 20th, 2018, 02:27 AM   #4
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Math Focus: Algebraic Number Theory, Arithmetic Geometry
Quote:
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: 502 Thanks: 280 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. Thanks from Alexis87 and Country Boy

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