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.