Is n4 - 1 divisible by 5 when n is not divisible by 5?

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Homework Help Overview

The problem involves determining whether the expression n4 - 1 is divisible by 5 when n is not divisible by 5. The context is rooted in number theory and modular arithmetic.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss proof by cases, considering different forms of n based on its divisibility by 5. Some suggest splitting cases further to explore various scenarios, while others propose using modular arithmetic and the binomial theorem to analyze the expression.

Discussion Status

The discussion is active, with participants exploring different representations of n and considering the implications of these forms on the divisibility of the expression. There is no explicit consensus yet, but several lines of reasoning are being examined.

Contextual Notes

Participants note the importance of considering various forms of n, specifically those that are not divisible by 5, to analyze the problem effectively. There is an emphasis on the need for careful consideration of assumptions and the structure of the proof.

GP35
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Homework Statement



Show n4 - 1 is divisible by 5 when n is not divisible by 5.

Homework Equations


The Attempt at a Solution



My thought here is proof by cases, one case being where n is divisible by 5 and the other case saying n is not divisible by 5. I am just not totally sure how to implement this strategy.
 
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If you split your second case a bit more that will do fine.
 
If n is not divisible by 5, then it is of the form n= 5k+ 1, n= 5k+ 2, n= 5k+ 3, or n= 5k+ 4.

Or you can consider n= 5k- 2, n= 5k- 1, n= 5k+ 1, and n= 5k+ 2 if that makes the calculations simpler.
 
HallsofIvy said:
If n is not divisible by 5, then it is of the form n= 5k+ 1, n= 5k+ 2, n= 5k+ 3, or n= 5k+ 4.

Or you can consider n= 5k- 2, n= 5k- 1, n= 5k+ 1, and n= 5k+ 2 if that makes the calculations simpler.

If we would to spent a little more time to think through , we will realize that the determining factor will then lies in [tex]5k \pm m[/tex] such the [tex]m^{4}\equiv 1 (mod5)[/tex] with application with binomial theorem.
 

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