Find all positive integers $n>1$ such that 
\[\tau(n)+\phi(n)=n+1\]
Which in this case, $\tau(n)$ represents the amount of positive divisors of $n$, and $\phi(n)$ represents the amount of positive integers which are less than $n$ and relatively prime with $n$.

[i]Raja Oktovin, Pekanbaru[/i]
To find all positive integers \( n > 1 \) such that 
\[
\tau(n) + \phi(n) = n + 1,
\]
we need to analyze the properties of the divisor function \(\tau(n)\) and the Euler's totient function \(\phi(n)\).

1. **Case \( n = p \) where \( p \) is a prime:**

   - For a prime number \( p \), the number of divisors \(\tau(p)\) is 2 (since the divisors are 1 and \( p \)).
   - The Euler's totient function \(\phi(p)\) is \( p-1 \) (since all numbers less than \( p \) are relatively prime to \( p \)).
   
   Therefore, for \( n = p \):
   \[
   \tau(p) + \phi(p) = 2 + (p - 1) = p + 1.
   \]
   This satisfies the given equation \( \tau(n) + \phi(n) = n + 1 \).

2. **Case \( n = 4 \):**

   - For \( n = 4 \), the number of divisors \(\tau(4)\) is 3 (since the divisors are 1, 2, and 4).
   - The Euler's totient function \(\phi(4)\) is 2 (since the numbers less than 4 that are relatively prime to 4 are 1 and 3).
   
   Therefore, for \( n = 4 \):
   \[
   \tau(4) + \phi(4) = 3 + 2 = 5.
   \]
   This satisfies the given equation \( \tau(n) + \phi(n) = n + 1 \).

3. **Case \( n \neq 4 \) and \( n \neq p \):**

   - Suppose \( n \) is a composite number that is not equal to 4. Let \( n = pq \) where \( p \) and \( q \) are distinct primes.
   - The number of divisors \(\tau(n)\) for \( n = pq \) is 4 (since the divisors are 1, \( p \), \( q \), and \( pq \)).
   - The Euler's totient function \(\phi(n)\) for \( n = pq \) is \((p-1)(q-1)\) (since the numbers less than \( pq \) that are relatively prime to \( pq \) are those that are not divisible by either \( p \) or \( q \)).
   
   Therefore, for \( n = pq \):
   \[
   \tau(pq) + \phi(pq) = 4 + (p-1)(q-1).
   \]
   We need to check if this can ever equal \( pq + 1 \):
   \[
   4 + (p-1)(q-1) = pq + 1.
   \]
   Simplifying, we get:
   \[
   4 + pq - p - q + 1 = pq + 1,
   \]
   \[
   5 - p - q = 0,
   \]
   \[
   p + q = 5.
   \]
   The only pairs of primes that sum to 5 are \( (2, 3) \) and \( (3, 2) \). However, if \( p = 2 \) and \( q = 3 \), then \( n = 6 \):
   \[
   \tau(6) = 4 \quad \text{and} \quad \phi(6) = 2.
   \]
   \[
   \tau(6) + \phi(6) = 4 + 2 = 6 \neq 7.
   \]
   Thus, \( n = 6 \) does not satisfy the equation. Therefore, there are no other solutions.

Conclusion:
The only solutions are \( n = 4 \) and \( n = p \) where \( p \) is a prime.

The final answer is \( \boxed{ n = 4 } \) and \( n = p \) where \( p \) is a prime.