Find all positive integers $n$ satisfying the following conditions simultaneously:

(a) the number of positive divisors of $n$ is not a multiple of $8$;
(b) for all integers $x$, we have
\[x^n \equiv x \mod n.\]
[i]
Proposed by usjl[/i]
1. **Verification of the given solutions:**
   - For \( n = 1 \):
     - The number of positive divisors of \( n = 1 \) is 1, which is not a multiple of 8.
     - For all integers \( x \), \( x^1 \equiv x \pmod{1} \) holds trivially.
   - For \( n \) being a prime \( p \):
     - The number of positive divisors of \( p \) is 2, which is not a multiple of 8.
     - For all integers \( x \), by Fermat's Little Theorem, \( x^p \equiv x \pmod{p} \).

2. **Proving that these are the only solutions:**
   - Let \( n \) be a positive integer satisfying the given conditions.
   - **Step 1: Show that \( n \) is squarefree.**
     - Let \( p \) be a prime divisor of \( n \). Consider \( x = p \).
     - We have \( p^n \equiv p \pmod{n} \). This implies \( p(p^{n-1} - 1) \equiv 0 \pmod{n} \).
     - Since \( p \) divides \( n \), \( p \) must divide \( p^{n-1} - 1 \).
     - If \( p^2 \) divides \( n \), then \( p^2 \) divides \( p^{n-1} - 1 \), which is not possible because \( p^{n-1} - 1 \) is not divisible by \( p^2 \) (as \( p \) is a prime and \( n-1 \geq 1 \)).
     - Therefore, \( p^2 \nmid n \), and \( n \) is squarefree.

   - **Step 2: Determine the number of prime divisors of \( n \).**
     - Let \( n = p_1 p_2 \cdots p_k \) be the prime factorization of \( n \).
     - The number of positive divisors of \( n \) is \( 2^k \).
     - Given that \( 2^k \) is not a multiple of 8, \( k \) must be less than 3 (since \( 2^3 = 8 \)).
     - Therefore, \( k \leq 2 \).

   - **Step 3: Consider the cases for \( k = 1 \) and \( k = 2 \).**
     - If \( k = 1 \), then \( n \) is a prime number.
     - If \( k = 2 \), let \( n = pq \) where \( p \) and \( q \) are distinct primes.
     - For all integers \( x \), \( x^{pq} \equiv x \pmod{pq} \) must hold.
     - By the Chinese Remainder Theorem, this implies \( x^{pq} \equiv x \pmod{p} \) and \( x^{pq} \equiv x \pmod{q} \).
     - By Fermat's Little Theorem, \( x^p \equiv x \pmod{p} \) and \( x^q \equiv x \pmod{q} \).
     - However, \( x^{pq} \equiv x \pmod{p} \) and \( x^{pq} \equiv x \pmod{q} \) cannot hold for all \( x \) unless \( p = q \), which contradicts the assumption that \( p \) and \( q \) are distinct primes.
     - Therefore, \( n \) cannot have exactly 2 prime divisors.

3. **Conclusion:**
   - The only possible values for \( n \) are 1 and prime numbers.

The final answer is \( \boxed{ n = 1 } \) and all prime numbers.