Prove that 3^(2n+1) + 4^(2n+1)is divided by 7

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  • #1
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prove that 3^(2n+1) + 4^(2n+1)
is divided by 7 for every natural n
 

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  • #2
Zurtex
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What have you done so far?
 
  • #3
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this isn't that hard. It is just induction as you suspect. The only fact you need to know is that [tex]c_1n + c_2m \equiv b_1n + b_2m\, (mod\, k)[/tex] if [tex] c_1\equiv b_1,\,c_2\equiv b_2\,(mod\, k)[/tex].

So just write down your sequence [tex]a_n = 3^{2n+1}+4^{2n+1}[/tex]. When [tex]a_n[/tex] is in that form can you write down [tex]a_{n+1}[/tex] without out modifying the exponent.

Hope those hints help.
Steven
 
  • #4
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well assuming that 3^(2n+1) + 4^(2n+1) is divisible by 7
im writing down what needs to be proven for n=n+1

3^(2n+3) + 4^(2n+3) / 7 = some integer = Z

9*3^(2n+1)+16*4^(2n+1) / 7 = Z

3^(2n+1)+4^(2n+1) / 7 + 8*3^(2n+1)+15*4^(2n+1) / 7 = Z

the first term is divisible because of the induction assumption so that leaves me with
8*3^(2n+1)+15*4^(2n+1) / 7

here im stuck i can't find a way to show that this term is divisible by 7
i tried proving that with another induction but that led me no where.
i also tried transforming 8*3^(2n+1)+15*4^(2n+1) / 7 to
7*3^(2n+1)+7*4^(2n+1) / 7 + 3^(2n+1)+8*4^(2n+1) / 7

3^(2n+1)+4^(2n+1) + 3^(2n+1)+8*4^(2n+1) / 7
and that leaves me with the term 3^(2n+1)+8*4^(2n+1) / 7
that does not help me i can't find a way to show that its divisible by 7
any help/tips would be appreciated

sorry for not using latex by the way it would just take me too much time to type all this in latex :smile:
 
  • #5
127
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Oh you're so close!

(3^(2n+1)+4^(2n+1)) / 7 + (8*3^(2n+1)+15*4^(2n+1)) / 7
= (3^(2n+1)+4^(2n+1)) / 7 + (3^(2n+1)+4^(2n+1)) / 7 + (7*3^(2n+1)+14*4^(2n+1)) / 7

You just needed to do your trick one more time over

Hope that helps
Steven
 
  • #6
88
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of course! i can't believe i didn't notice this thank you very much! :smile:
 
  • #7
We can easily prove it without induction:

[tex]a_n = 3^{2n+1}+4^{2n+1}\equiv 3^{2n+1} + (-3)^{2n+1} \equiv 3^{2n+1}\cdot(1 + (-1)^{2n+1}) \equiv 0 (mod 7)[/tex].
 

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