Prove using nEN that n3+2 will never be divisible by 8

To prove that \( n^3 + 2 \) will never be divisible by 8 for any \( n \in \mathbb{N} \), we can examine the expression \( n^3 + 2 \) modulo 8.

### Step 1: Check the possible values of \( n \mod 8 \)
For any integer \( n \), there are 8 possible values for \( n \mod 8 \) (0, 1, 2, 3, 4, 5, 6, 7). We will compute \( n^3 + 2 \mod 8 \) for each of these values and verify if any of them can result in 0 (i.e., \( n^3 + 2 \equiv 0 \mod 8 \)).

### Step 2: Compute \( n^3 + 2 \mod 8 \) for each \( n \mod 8 \)
- **If \( n \equiv 0 \mod 8 \):**
  \[
  n^3 + 2 = 0^3 + 2 = 2 \equiv 2 \mod 8
  \]
  
- **If \( n \equiv 1 \mod 8 \):**
  \[
  n^3 + 2 = 1^3 + 2 = 1 + 2 = 3 \equiv 3 \mod 8
  \]
  
- **If \( n \equiv 2 \mod 8 \):**
  \[
  n^3 + 2 = 2^3 + 2 = 8 + 2 = 10 \equiv 2 \mod 8
  \]
  
- **If \( n \equiv 3 \mod 8 \):**
  \[
  n^3 + 2 = 3^3 + 2 = 27 + 2 = 29 \equiv 5 \mod 8
  \]
  
- **If \( n \equiv 4 \mod 8 \):**
  \[
  n^3 + 2 = 4^3 + 2 = 64 + 2 = 66 \equiv 2 \mod 8
  \]
  
- **If \( n \equiv 5 \mod 8 \):**
  \[
  n^3 + 2 = 5^3 + 2 = 125 + 2 = 127 \equiv 7 \mod 8
  \]
  
- **If \( n \equiv 6 \mod 8 \):**
  \[
  n^3 + 2 = 6^3 + 2 = 216 + 2 = 218 \equiv 2 \mod 8
  \]
  
- **If \( n \equiv 7 \mod 8 \):**
  \[
  n^3 + 2 = 7^3 + 2 = 343 + 2 = 345 \equiv 1 \mod 8
  \]

### Step 3: Conclusion
From the computations above, we see that for all possible values of \( n \mod 8 \), the expression \( n^3 + 2 \) is never congruent to 0 modulo 8. Specifically, the possible remainders are 1, 2, 3, 5, or 7, but never 0.

Therefore, \( n^3 + 2 \) is never divisible by 8 for any \( n \in \mathbb{N} \).

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Would you like any further details or have any questions about this proof?

Here are 5 questions related to this:

1. Can you prove a similar statement for \( n^3 - 2 \)?
2. What happens if you check \( n^3 + 3 \) modulo 8?
3. Can you generalize this type of proof to other moduli, such as 4 or 16?
4. How does this proof change if \( n \in \mathbb{Z} \) instead of \( n \in \mathbb{N} \)?
5. Can \( n^3 + 2 \) ever be divisible by a prime greater than 8?

**Tip**: When working with modular arithmetic, computing small powers modulo a number often helps in spotting patterns for proofs.

Prove using nEN that n^3 + 2 will never be divisible by 8

Explore a proof using modular arithmetic to show that n^3 + 2 will never be divisible by 8 for any natural number n. The problem involves examining the expression n^3 + 2 modulo 8 and determining the possible remainders. Suitable for high school students in Grades 10-12.

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