Divisibility problem cant use module just induction

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Willy_Will
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Hi all,

Great forum, I have been reading some cool stuff here for about a month.

Heres my question:
Using induction prove that 5 divides 8^n - 3^n, n Natural Number.

I know its true for n = 1, but I get stuck on the n = k+1. I don't know how to proceed from here: 8(8^k) - 3(3^k)

Also, I KNOW this is very easy using mod 5 but I can't do it here, I HAVE to prove it using only induction.

Any hints?

Thanks,

-William
 
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I think so, but how can I use that in my problem?
 
I think I've got it.

Suppose that 5 divides 8^k - 3^k ---> 8^k - 3^k = 5x, for some integer x.

case n = k +1
= 8^k+1 - 3^k+1
= 8(8^k) - 3(3^k)
= 8(8^k) - (3^k)(8) + (3^k)(8) - (3)(3^k)
= 8(8^k - 3^k) + 3^k(8 - 3)
(1) (2)

(1) by hypotesis 5 divides the expresion.
(2) 5 = 8 - 3, 5 divides 5.

= 8(5x) + 3^k(5)
= 5(8x + 3^k)

5 divides the expresion

Please tell me its that way, if not, point the errors and some hints.

Thanks, really.

-William
 
Last edited:
I think I've got it.

Suppose that 5 divides 8^k - 3^k ---> 8^k - 3^k = 5x, for some integer x.

case n = k +1
= 8^k+1 - 3^k+1
= 8(8^k) - 3(3^k)
= 8(8^k) - (3^k)(8) + (3^k)(8) - (3)(3^k)
= 8(8^k - 3^k) + 3^k(8 - 3)
(1) (2)

(1) by hypotesis 5 divides the expresion.
(2) 5 = 8 - 3, 5 divides 5.

= 8(5x) + 3^k(5)
= 5(8x + 3^k)

5 divides the expresion

Please tell me its that way, if not, point the errors and some hints.

Thanks, really.

-William
 
sorry for the double post.

(1) = (8^k - 3^k)
(2) = (8 - 3)

(1) by hypotesis 5 divides the expresion.
(2) 5 = 8 - 3, 5 divides 5.

= 8(5x) + 3^k(5)
= 5(8x + 3^k)

5 divides the expresion
 
Thanks guys, I really appreciate it!