To solve this problem, let's analyze each condition and look for the smallest integer $n$ that satisfies both.

Step 1: First Condition Analysis

We are given: $n^2 \equiv 1 \pmod{7}$ This condition means that $n^2$ leaves a remainder of 1 when divided by 7. The possible values of $n$ modulo 7 that satisfy $n^2 \equiv 1 \pmod{7}$ are typically the solutions to this congruence. Testing values from 0 to 6 modulo 7:

• $n \equiv 1 \pmod{7} \Rightarrow n^2 \equiv 1^2 = 1 \pmod{7}$
• $n \equiv 6 \pmod{7} \Rightarrow n^2 \equiv 6^2 = 36 \equiv 1 \pmod{7}$

So, $n \equiv 1$ or $n \equiv 6 \pmod{7}$ satisfies $n^2 \equiv 1 \pmod{7}$.

Step 2: Second Condition Analysis

We are also given: $n^3 \equiv 8 \pmod{11}$ We need to find the values of $n$ modulo 11 such that $n^3 \equiv 8 \pmod{11}$. Testing values from 0 to 10 modulo 11:

• $n \equiv 2 \pmod{11} \Rightarrow n^3 \equiv 2^3 = 8 \pmod{11}$

Thus, $n \equiv 2 \pmod{11}$ is the solution to $n^3 \equiv 8 \pmod{11}$.

Step 3: Using the Chinese Remainder Theorem

We now have two simultaneous congruences to solve:

1. $n \equiv 1$ or $n \equiv 6 \pmod{7}$
2. $n \equiv 2 \pmod{11}$

Let’s first try $n \equiv 1 \pmod{7}$:

1. $n = 7k + 1$ for some integer $k$.
2. Substitute into the second congruence: $7k + 1 \equiv 2 \pmod{11}$.
3. Simplify this: $7k \equiv 1 \pmod{11}$.

To solve $7k \equiv 1 \pmod{11}$, we find the multiplicative inverse of 7 modulo 11. The inverse is 8 (since $7 \times 8 = 56 \equiv 1 \pmod{11}$), so: $k \equiv 8 \pmod{11}$ Thus, $k = 11m + 8$ for some integer $m$, and substituting back: $n = 7(11m + 8) + 1 = 77m + 56 + 1 = 77m + 57$ The smallest positive $n$ is obtained by setting $m = 0$, giving $n = 57$.

Verification

1. $n = 57$:
   • $57 \mod 7 = 1$, so $n^2 \equiv 1 \pmod{7}$ holds.
   • $57 \mod 11 = 2$, so $n^3 \equiv 8 \pmod{11}$ also holds.

Therefore, the smallest positive integer $n$ that satisfies both conditions is: $\boxed{57}$

Would you like further details or have any questions?

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Related Questions

1. What if the modulus for the first condition was different, such as 5 instead of 7?
2. How would we approach this if we needed the smallest negative integer $n$?
3. Can we generalize this method to other powers or moduli?
4. How does the Chinese Remainder Theorem work for more than two simultaneous congruences?
5. What other techniques are available for finding multiplicative inverses in modular arithmetic?

Tip: The Chinese Remainder Theorem is a powerful tool for solving congruences and is especially useful when the moduli are coprime.