Question

Hmm.....how about a little shake up for once?

Prove that 8^k-5^k is divisible by 3.

Answer 1

I can see two ways to solve this problem. I'll award medals for each :D

Answer 2

1. the difference in the coefficient is 3 so that is a clue right there :)

Answer 3

Lol I know. I just wanna see how people approach this problem. I'm not searching for answers or anything.

Answer 4

\[\lim_{k\rightarrow\infty}(\frac{8^k-5^k}{3})=\infty\]
thus
\[\frac{8^k-5^k}{3} ~~k\in \mathbb{R} ~~\forall:k\]

Answer 5

hmm........You didn't exactly prove my challenge.

Answer 6

your question isn't as well defined as you probably need it to be. You dont say that k must be an integer, nor real. You also dont say it needs to be evenly divisible by 3 (have an integer solution). Truthfully, the best proof to what you've asked is this:

i. The postulate of addition
a+b=c
ii. Multiplication is a trivial section of addition
a+a=a*a
iii. Exponentiation is a trivial section of multiplication
a+a=a*a=a^2
iv. Division is a trivial section of exponentiation
a*a^(-1)=1

thus all numbers are divisible by 3, since all numbers are able to be added (i).

Answer 7

on ii. I meant,
a+a+a+...+a (where there are "a" components of the sum) = a*a

Answer 8

Ok, so in this case, let me rewrite this:

Prove the following:

For \[k \ge 0, k \epsilon \mathbb{Z}^+\],
\[8^k -5^k\], is divisible by three for all of k.

Answer 9

and no agreene, unfortunately that was not what I meant.

Answer 10

Prove
\[\frac{8^k-5^k}{3}=c ~~~ c\in\mathbb{Z} ~~and~~ k\in \mathbb{Z} \forall :{c,k}\]

I think that's the question you actually want to ask.

Answer 11

^yep.

Answer 12

Well, I'm a little late to the party here, but I couldn't resist responding. I can immediately think of two ways I might show this.

Easy way: Look at \(8^k-5^k \pmod{3}\). Since \(8^k\equiv 2^k\pmod3\) and likewise with \(5^k\) the problem reduces to \(2^k-2^k\equiv0\pmod3\) so it is divisible by 3.

Harder way for those not familiar with modular arithmetic on prime numbers: Write \[8^k-5^k=(6+2)^k-(3+2)^k\]Let's look at the expansion of \((6+2)^k\). Note that the only term that won't be divisible by 3, is the \(2^k\) term. Hence, \((6+2)^k=3a+2^k\) for some \(a\). From a similar argument, \((3+2)^k=3b+2^k\) for some \(b\). It follows that \[8^k-5^k=3a+2^k-3b-2^k=3(a-b)\]So we see that it is divisible by 3.

Answer 13

@KingGeorge LOL I forgot I haven't closed this. And very interesting proofs. I think when I made this post, I didn't have in mind modular proofs. I think I had thought up these options:
1) Proof by induction, which is straightforward: Assume that \[8^k - 5^k = 3k, k \epsilon Z\] . P(1) is true, so assume P(k). Finding P(k+1):\[8^(k+1) - 5^(k+1)= 8*8^k - 5*5^k = 3*8^k +5(8^k - 5^k) = 3*8^k + 5(3n) = 3(8^k +5n), thus proving by induction the following.\]
2) Using a little formula: \[a^n - b^n = (a - b)(a^n-1 +a^n-2*b.............+b^n)\], and substituting with 8 and 5 for a and b respectively.

Answer 14

I should have thought of the induction proof. I've always liked induction. Anyways, great problem.