- #1

Anzas

- 87

- 0

prove that 3^(2n+1) + 4^(2n+1)

is divided by 7 for every natural n

is divided by 7 for every natural n

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- Thread starter Anzas
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- #1

Anzas

- 87

- 0

prove that 3^(2n+1) + 4^(2n+1)

is divided by 7 for every natural n

is divided by 7 for every natural n

- #2

Zurtex

Science Advisor

Homework Helper

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What have you done so far?

- #3

snoble

- 127

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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

Anzas

- 87

- 0

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

- #5

snoble

- 127

- 0

(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

Anzas

- 87

- 0

of course! i can't believe i didn't notice this thank you very much!

- #7

Algebracus

- 2

- 0

[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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