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

  • Thread starter Thread starter GP35
  • Start date Start date
  • Tags Tags
    Proof
AI Thread Summary
The discussion centers on proving that n^4 - 1 is divisible by 5 when n is not divisible by 5. Participants suggest using proof by cases, specifically considering values of n in the forms 5k + 1, 5k + 2, 5k + 3, and 5k + 4. Alternative forms like 5k - 2 and 5k - 1 are also proposed for simplification. The key insight is that the divisibility hinges on the expression m^4 being congruent to 1 modulo 5, which can be examined using the binomial theorem. This approach aims to clarify the conditions under which the original statement holds true.
GP35
Messages
1
Reaction score
0

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.
 
Physics news on Phys.org
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 5k \pm m such the m^{4}\equiv 1 (mod5) with application with binomial theorem.
 

Thread 'Finding the number of ways to arrange identical balls in a circle (3 different colors)'

I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
View full post »

Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
View full post »

Similar threads

Proof That an Integer is Divisible by 2
Replies
3
Views
2K
Prove that ##5 | (2^n a) \rightarrow (5 | a)##for any ##n##
Replies
10
Views
3K
Prove that the product of 4 consecutive numbers cannot be a perfect square
Replies
14
Views
4K
Prove divisibility, mathematical induction
Replies
8
Views
2K
Prime numbers and divisibility by 12
Replies
5
Views
2K
Is This Proof That 8 Divides n^2 - 1 for Odd n Valid?
Replies
11
Views
5K
Polynomial Long Division for Limit Calculation
Replies
4
Views
1K
Prove that the product of any three consecutive integers is
Replies
11
Views
3K
Prove that {x∈ℤ : 2|x} ∩ {x∈ℤ : 9|x} ⊂ {x∈ℤ : 6|x}
Replies
17
Views
3K
A problem concerning divisibility and the number 31. (Number theory)
Replies
2
Views
2K

Hot Threads

Recent Insights

Back
Top