Prove 11^n - 4^n Divisible by 7 for All Natural n

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Homework Help Overview

The problem involves proving by induction that the expression 11^n - 4^n is divisible by 7 for all natural numbers n. The context is rooted in mathematical induction and number theory.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to establish the base case for n=1 and then assumes the statement holds for n=k, seeking guidance on how to proceed with n=k+1. Some participants suggest considering cases based on the parity of k+1, while others discuss factoring techniques related to differences of powers.

Discussion Status

Participants are actively exploring different approaches to the induction proof. Some guidance has been offered regarding factoring and the implications of substituting values into the expression. There is an ongoing exchange of ideas without a clear consensus on the best method yet.

Contextual Notes

Participants are navigating the constraints of the problem, including the requirement to use mathematical induction and the need to demonstrate divisibility by 7. There is mention of external resources that may provide additional insights, though their relevance is not fully clarified.

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Homework Statement



Prove by induction on n that [tex]11^n - 4^n[/tex] is divisible by 7 for all natural numbers n (not including zero).

Homework Equations



Principle of mathematical induction.

The Attempt at a Solution



Of course I start by showing that the statement holds for n=1

[tex]11^1 - 4^1 = 7[/tex]

and

[tex]7/7 = 1 \in \mathbb{Z}[/tex]

Then I go on to assume that the statement holds for some value k, which gives

[tex]\frac{11^k - 4^k}{7} = a \in \mathbb{Z}[/tex]

then I examine the situation for k+1

[tex]\frac{11^{k+1} - 4^{k+1}}{7}[/tex]

and I must show that this is also equal to an integer.

However I have tried rewriting this in many ways in order that I may be able to use the induction assumption. But I have had no luck. Any hints to help me from here?

Thanks.
 
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There are two cases here:

Case 1: k+1 even

Case 2: k+1 odd

What do you think you can do if k+1 is even? I'm still thinking about the odd case.

edit:

found it: http://www.k12math.com/math-concepts/algebra/factoring/factoring.htm
 
Last edited by a moderator:
gb7nash said:
found it: http://www.k12math.com/math-concepts/algebra/factoring/factoring.htm

Thank you. But what part of this page are you referring to? The bottom part about factoring differences of odd powers?
 
Last edited by a moderator:
AlexChandler said:
Thank you. But what part of this page are you referring to? The bottom part about factoring differences of odd powers?

You got it.
 
[tex] 11^{n+1} - 4^{n+1} = 11 \cdot 11^{n} - 4 \cdot 4^{n}[/tex]

Right? Now, remember that

[tex] 11 = 7 + 4[/tex]

Plug that in for your first 11 and see what happens. : )
 
And to be more specific, whenever you have a difference of the same odd powers an-bn, you can always factor it into:

(a-b)(an-1+an-2b+an-3b2+...+bn-1)

l'Hôpital said:
[tex] 11^{n+1} - 4^{n+1} = 11 \cdot 11^{n} - 4 \cdot 4^{n}[/tex]

Right? Now, remember that

[tex] 11 = 7 + 4[/tex]

Plug that in for your first 11 and see what happens. : )

That's slick. I didn't see that. You could do it either way, but I would suggest this method. It's a lot shorter.
 
l'Hôpital said:
[tex] 11^{n+1} - 4^{n+1} = 11 \cdot 11^{n} - 4 \cdot 4^{n}[/tex]

Right? Now, remember that

[tex] 11 = 7 + 4[/tex]

Plug that in for your first 11 and see what happens. : )

Aha! Thank you. I had a feeling it would be something simple like this.

Thank you both very much for your help.
 

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