Hello there!
I don't get it - what to assume in the following problem for solving by induction
Prove that 6 divides n^3 - n whenever n is a nonnegative integer
I tried following:
assume for "k" --
k^3 - k = 6r for some integer "r"
now to prove "k+1" we get:
(k+1)^3 - (k+1) = k^3 + 3k^2+3k+1 - k - 1
= (k^3-k) +3k^2+3k
= 6r + 3k^2+3k by induction
what should i do after this? I am not getting "6" as a common factor how come 6 divides n^3-n!!!
Huh. I didn't see that. Well you could start by writing which is automatically divisible by 6 and not worry about it. I guess this problem is just lousy for an induction proof.
But we should be able to write it as one.
Assume the argument is true for some n = k. Then we have
We need to show that
So
See what you can do with that line.
-Dan
EDIT: NVM, you addressed this while I was writing it.
But is that legitimate? Because we could also say
Now, if we are required to use induction to prove is divisible by 6, then why are we not required to use induction when we prove it for ?
It seems like circumventing the requirements.