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 August 26th, 2012, 08:45 PM #1 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Modular Arithmetic Can anyone shed a little light on this problem? Prove that if $k \in \mathbb{N}$ is a multiple of 4 then $7^{4n} \equiv \hspace{1mm} _5 \hspace{1mm} 1$ It may help to know that $7^4=2,401$ The solution I was given was to use induction
 August 26th, 2012, 09:11 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,164 Thanks: 472 Math Focus: Calculus/ODEs Re: Modular Arithmetic I'm not sure about your notation, but I am guessing you are to prove that: $7^{4n}=5k+1$ where $k,n\in\mathbb{N}$ Is this correct?
August 27th, 2012, 08:34 AM   #3
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Re: Modular Arithmetic

Hello, aaron-math!

Quote:
 $\text{Prove that if }k \in \mathbb{N} \text{ is a multiple of 4 then }7^{4n}\:\equiv\:1\text{ (mod 5)}$ $\text{It may help to know that }7^4\,=\,2,401$ $\text{The solution I was given was to use induction}$

$\text{Verify }S(1):\;\;7^4 \:=\:2401 \;\equiv\;1\text{ (mod 5) \; . . . \;True!}$

$\text{Assume }S(k):\;\;7^{4k} \:\equiv\:1\text{ (mod 5)}$

$\text{Consider }S(k+1):\;\;7^{4(k+1)} \:=\:7^{4k+4} \:=\:7^4\,\cdot\,7^k$

$\text{Since }\,7^4\:\equiv\;1\text{ (mod 5)}\,\text{ and }\,7^k \:\equiv\:1\text{ (mod 5)}
\;\;\;\text{we have: }\:7^{4(k+1)} \;\equiv\;1\,\cdot\,1\text{ (mod 5)} \;$

$\text{Therefore: }\:7^{4(k+1)} \;\equiv\;1\text{ (mod 5)}$

$\text{The inductive proof is complete.}$

 August 27th, 2012, 08:51 AM #4 Senior Member   Joined: Mar 2012 Posts: 572 Thanks: 26 Re: Modular Arithmetic A folksier version: Any number that ends in 1 = 1 mod 10 = 1 mod 5 Any number that ends in 1 x any number that ends in 7 ends in 7. (7^1, 7^5, 7^9...) Any number that ends in 7 x any number that ends in 7 ends in 9. (7^2, 7^6, 7^10...) Any number that ends in 9 x any number that ends in 7 ends in 3. (7^3, 7^7, 7^11...) Any number that ends in 3 x any number that ends in 7 ends in 1. (7^4, 7^8, 7^12...)

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