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I think you meant to say `Induction Hypothesis: 8^k-3^k is divisible by 5`

if we assume that claim is true, then we can say
8^k-3^k = 5q
where q is some integer. Agreed?

what is the significance of 5q??

it just means "5 times some integer"

if I said "100 is divisible by 5" I can state that as 100 = 5q
In this case, q = 20

ok, yeah. so that just means divisible by 5

yeah

i do not understand.8^(k+1) is just 64, no? not 64^K?

how are you getting 64? you can't replace k with 1

i didn't know i couldn't do that.

do you see how I turned `8^(k+1)-3^(k+1)` into `8*8^k-3*3^k` ?

no, that was my next question. i thought k+1 would turn into ^2

recall that
\[\LARGE x^{y+z} = x^y * x^z\]

so \[\Large 8^{k+1} = 8^k*8^1 = 8^k*8 = 8*8^k\]
make sense?

yes. thank you. and now i have to do the same thing for the other side, right?

for the 3^(k+1) part you mean? yes

ok, so i just get 3^k*3+5q=6k+5q

multiply

40q + 24^k - 9^k
40q + 15^k

8*3^k isn't equal to 24^k

im going to have to study that, tomorrow. can you look at a different kind of problem with me?

sure

1^3 + 2^3 + 3^3 + … + n^3 = n^2(n + 1)^2/4 for all positive integers n

I'm sure you have the base case already done?

no. i just started looking at it. i have not tried solving a sequence kind of problem yet.

n^2(n^2+1)/4

you'll use induction here

i have less idea about this problem then i did the last :|

the base case is n = 1

prove that is true