Let's tackle these proofs separately.

Part (a): If $8 \nmid (n^2 - 1)$, then $n$ is even.

Proof:

1. Consider the possible values of $n$ modulo $8$: For any integer $n$, $n$ modulo $8$ can take values $0, 1, 2, 3, 4, 5, 6,$ or $7$.

2. Calculate $n^2 - 1$ modulo $8$ for each case:

   • If $n \equiv 0 \pmod{8}$, then $n^2 - 1 \equiv 0^2 - 1 \equiv -1 \equiv 7 \pmod{8}$.
   • If $n \equiv 1 \pmod{8}$, then $n^2 - 1 \equiv 1^2 - 1 \equiv 0 \pmod{8}$.
   • If $n \equiv 2 \pmod{8}$, then $n^2 - 1 \equiv 2^2 - 1 \equiv 4 - 1 \equiv 3 \pmod{8}$.
   • If $n \equiv 3 \pmod{8}$, then $n^2 - 1 \equiv 3^2 - 1 \equiv 9 - 1 \equiv 8 \equiv 0 \pmod{8}$.
   • If $n \equiv 4 \pmod{8}$, then $n^2 - 1 \equiv 4^2 - 1 \equiv 16 - 1 \equiv 15 \equiv 7 \pmod{8}$.
   • If $n \equiv 5 \pmod{8}$, then $n^2 - 1 \equiv 5^2 - 1 \equiv 25 - 1 \equiv 24 \equiv 0 \pmod{8}$.
   • If $n \equiv 6 \pmod{8}$, then $n^2 - 1 \equiv 6^2 - 1 \equiv 36 - 1 \equiv 35 \equiv 3 \pmod{8}$.
   • If $n \equiv 7 \pmod{8}$, then $n^2 - 1 \equiv 7^2 - 1 \equiv 49 - 1 \equiv 48 \equiv 0 \pmod{8}$.

3. Identify when $8 \nmid (n^2 - 1)$:

   • We see that $n^2 - 1 \equiv 0 \pmod{8}$ if $n \equiv 1$, $3$, $5$, or $7$ modulo $8$, which correspond to odd values of $n$.
   • Conversely, when $n \equiv 0$, $2$, $4$, or $6$ modulo $8$, $n$ is even, and $8 \nmid (n^2 - 1)$.

Therefore, if $8 \nmid (n^2 - 1)$, then $n$ must be even.

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Part (b): If $3 \nmid (n^2 - 1)$, then $3 \mid n$.

Proof:

1. Consider the possible values of $n$ modulo $3$: For any integer $n$, $n$ modulo $3$ can take values $0$, $1$, or $2$.

2. Calculate $n^2 - 1$ modulo $3$ for each case:

   • If $n \equiv 0 \pmod{3}$, then $n^2 - 1 \equiv 0^2 - 1 \equiv -1 \equiv 2 \pmod{3}$.
   • If $n \equiv 1 \pmod{3}$, then $n^2 - 1 \equiv 1^2 - 1 \equiv 0 \pmod{3}$.
   • If $n \equiv 2 \pmod{3}$, then $n^2 - 1 \equiv 2^2 - 1 \equiv 4 - 1 \equiv 3 \equiv 0 \pmod{3}$.

3. Identify when $3 \nmid (n^2 - 1)$:

   • We see that $n^2 - 1 \equiv 0 \pmod{3}$ if $n \equiv 1$ or $2$ modulo $3$, which correspond to cases where $3$ does not divide $n$.
   • Conversely, when $n \equiv 0 \pmod{3}$, $n$ is divisible by $3$, and $3 \nmid (n^2 - 1)$.

Therefore, if $3 \nmid (n^2 - 1)$, then $3 \mid n$.

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Would you like further details or have any questions?

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Additional Questions

1. What are alternative methods to prove part (a)?
2. How can we extend this proof for other primes besides $3$ and $8$?
3. What does this proof suggest about divisibility rules for powers of integers?
4. Can modular arithmetic properties be used to prove similar divisibility conditions?
5. How does this proof relate to congruences in number theory?

Tip

Understanding proofs by exhaustion for modular cases can simplify divisibility proofs, especially with small moduli like $8$ or $3$.