Mathematica Prove by Mathematical Induction: (13^n)-(6^n) Divisible by 7

AI Thread Summary
The discussion focuses on proving by mathematical induction that (13^n) - (6^n) is divisible by 7. Participants outline the base case and the inductive step, where they assume the statement holds for n = k and aim to prove it for n = k + 1. They explore the factorization of (13^(k+1)) - (6^(k+1)) and how to express it in terms of (13^k) - (6^k). A suggestion is made to utilize a standard factorization for a^n - b^n to simplify the proof. The conversation highlights the importance of recognizing patterns in mathematical induction for divisibility proofs.
supasupa
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Hey there evryone

I need some help with this problem as I don't know which direction to go with it.

Prove by mathematical induction that (13^n)-(6^n) is divisible by 7.



The Base Step is obviously ok...

Then assume (13^K)-(6^K) is true

Then have to prove (13^(k+1))-(6^(k+1)) is true... how do i do this

thank heaps...
 
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Can you factor (13^(k+1))-(6^(k+1)) ?
 
yeah that's what i have been trying to do but i don't know how to get

(13^(k+1) - 6^(k+1))

as a multiple of (13^k - 6^k).

I don't know if this is the correct direction to take
 
13^{k+1}-6^{k+1}=13 \cdot (13^k-6^k)+7 \cdot 6^k
 
how do you do that?
like what is the step u take to get there
 
i get

13.13^K - 6.6^K

then what do i do??
 
Instead of trying that... can you find a way to just show (13^(k+1) - 6^(k+1)) is divisible by seven? What looks like it equal seven in that equation?
 
supasupa said:
i get

13.13^K - 6.6^K

then what do i do??

=13.13^K-(13-7).6^K=13(13^K-6^K)+7.6^K
 
thank you very much...that makes a heap more sense now
:smile:
 
  • #10
In all this, you've missed out the slightly easier solution Shredder was trying to point you to. There is a standard factorization for a^n - b^n. You use a specific case of this factorization when you write the sum of a geometric series : 1+b+b^2+...+b^{n-1} = (1-b^n)/(1-b)
 
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