Which cube members are not in the sequence and prove it?

Natasha1
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Which cube members are not in the sequence and prove it?

2, 5, 8, 11, 14, ...

How can this be proved :cry:

My answer:

an = 3n + 2

Any natural number may be written as N= 3k+p for some natural number K and p=0,1 or 2.

So

N^3=(3k+p)3
N^3=3(9k+k^2p+kp^2)+p^3
N^3=3k+p^3

Therefore N^3 (mod 3)

So as an is congruent to 2(mod 3), N^3 lies in the sequence. In other words, the cude of any number in the sequence is in the sequence. But how do I go and show which cube members are not in the sequence and moreover how do I prove it?
 
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You want to prove something of the form:

a cube x³ is not in the sequence if and only if ________

where you fill in the blank with something informative. You have to understand that about the question, that it is somewhat open-ended. Technically, you could fill in the blank with "x³ is not in the sequence" or "x³ is not congruent to 2 (mod 3)" but neither answer is very informative. I'm actually pretty stumped as to how to explain how to do this. If you start with the condition "x³ is not in the sequence," will you be able to churn out equivalent conditions such that you can prove:

"x³ is not in the sequence" if and only if A
A iff B
B iff C
C iff D

Until you reach something (in this case, D) which is informative, making your final answer "x³ is not in the sequence iff D". Or should you guess the answer, D, first, then work your way to finding sentences A, B, and C such that you can prove:

"x³ is not in the seq." iff A
A iff B
B iff C
C iff D?

Well, here are a few logical tautologies which you should understand (they hold for any sentences A, B, and C):

"A iff B" is equivalent to "A implies B and B implies A"
"A implies B" is equivalent to "not-B implies not-A"
"A iff B" is equivalent to "A implies B and not-A implies not-B" (putting the above two together)
"A iff B" is equivalent to "either both A and B are true, or both A and B are false - that is, A and B have the same truth-value, whatever that truth value may be"
A iff A
"A iff B, and B iff C" implies "A iff C", however
"A iff C" does not imply "A iff B and B iff C", so
"A iff C" is not equivalent to "A iff B, and B iff C"
"A iff B" is equivalent to "B iff A"
"A implies B" is NOT equivalent to "B implies A"
"A iff B" is equivalent to "not-A iff not-B"
 
So here are some good things to know:

"x³ is not in the sequence iff A" is equivalent to "x³ IS in the sequence iff NOT-A"

And

"any number n is in the sequence iff n = 2 (mod 3)"

So, in particular

"a cube, n = x³ is in the sequence iff x³ = 2 (mod 3)"
 
Natasha,

Here is a hint to get you going: it is helpful to write your candiate number as N = 3k + r where r = 0, \, 1, \, 2. You also correctly realized that when looking at N^3, all the factors containing k are already divisible by 3. Now, what you want is to have N^3 = 3 n + 2 for some n, right? What does this say about the one factor in N^3 that isn't explicitly divisible by 3?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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