For all positive integers $m,n$ define $f(m,n) = m^{3^{4n}+6} - m^{3^{4n}+4} - m^5 + m^3$. Find all numbers $n$ with the property that $f(m, n)$ is divisible by 1992 for every $m$. 

[i]Bulgaria[/i]
To solve the problem, we need to determine all positive integers \( n \) such that \( f(m, n) = m^{3^{4n}+6} - m^{3^{4n}+4} - m^5 + m^3 \) is divisible by 1992 for every \( m \). 

First, we factorize 1992:
\[ 1992 = 8 \times 3 \times 83 \]

We need to show that \( f(m, n) \) is divisible by 8, 3, and 83 for all \( m \).

### Divisibility by 3:
Consider \( f(m, n) \mod 3 \):
\[ f(m, n) = m^{3^{4n}+6} - m^{3^{4n}+4} - m^5 + m^3 \]

If \( m \equiv 0 \mod 3 \), then clearly \( f(m, n) \equiv 0 \mod 3 \).

If \( m \not\equiv 0 \mod 3 \), then \( m \equiv 1 \mod 3 \) or \( m \equiv 2 \mod 3 \). We need to check both cases:
- If \( m \equiv 1 \mod 3 \):
  \[ m^{3^{4n}+6} \equiv 1 \mod 3 \]
  \[ m^{3^{4n}+4} \equiv 1 \mod 3 \]
  \[ m^5 \equiv 1 \mod 3 \]
  \[ m^3 \equiv 1 \mod 3 \]
  Thus,
  \[ f(m, n) \equiv 1 - 1 - 1 + 1 \equiv 0 \mod 3 \]

- If \( m \equiv 2 \mod 3 \):
  \[ m^{3^{4n}+6} \equiv 2^{3^{4n}+6} \equiv (-1)^{3^{4n}+6} \equiv -1 \equiv 2 \mod 3 \]
  \[ m^{3^{4n}+4} \equiv 2^{3^{4n}+4} \equiv (-1)^{3^{4n}+4} \equiv 1 \mod 3 \]
  \[ m^5 \equiv 2^5 \equiv -1 \equiv 2 \mod 3 \]
  \[ m^3 \equiv 2^3 \equiv -1 \equiv 2 \mod 3 \]
  Thus,
  \[ f(m, n) \equiv 2 - 1 - 2 + 2 \equiv 1 \mod 3 \]

Since \( f(m, n) \equiv 0 \mod 3 \) in both cases, \( f(m, n) \) is divisible by 3 for all \( m \).

### Divisibility by 8:
Consider \( f(m, n) \mod 8 \):
\[ f(m, n) = m^{3^{4n}+6} - m^{3^{4n}+4} - m^5 + m^3 \]

If \( m \equiv 0 \mod 8 \), then clearly \( f(m, n) \equiv 0 \mod 8 \).

If \( m \not\equiv 0 \mod 8 \), then \( m \equiv 1 \mod 8 \) or \( m \equiv -1 \mod 8 \). We need to check both cases:
- If \( m \equiv 1 \mod 8 \):
  \[ m^{3^{4n}+6} \equiv 1 \mod 8 \]
  \[ m^{3^{4n}+4} \equiv 1 \mod 8 \]
  \[ m^5 \equiv 1 \mod 8 \]
  \[ m^3 \equiv 1 \mod 8 \]
  Thus,
  \[ f(m, n) \equiv 1 - 1 - 1 + 1 \equiv 0 \mod 8 \]

- If \( m \equiv -1 \mod 8 \):
  \[ m^{3^{4n}+6} \equiv (-1)^{3^{4n}+6} \equiv 1 \mod 8 \]
  \[ m^{3^{4n}+4} \equiv (-1)^{3^{4n}+4} \equiv 1 \mod 8 \]
  \[ m^5 \equiv (-1)^5 \equiv -1 \mod 8 \]
  \[ m^3 \equiv (-1)^3 \equiv -1 \mod 8 \]
  Thus,
  \[ f(m, n) \equiv 1 - 1 - (-1) + (-1) \equiv 0 \mod 8 \]

Since \( f(m, n) \equiv 0 \mod 8 \) in both cases, \( f(m, n) \) is divisible by 8 for all \( m \).

### Divisibility by 83:
Consider \( f(m, n) \mod 83 \):
\[ f(m, n) = m^{3^{4n}+6} - m^{3^{4n}+4} - m^5 + m^3 \]

If \( 83 \mid m \), then clearly \( f(m, n) \equiv 0 \mod 83 \).

If \( 83 \nmid m \), we need to show that \( f(m, n) \equiv 0 \mod 83 \). We need:
\[ m^{3^{4n}+6} - m^{3^{4n}+4} - m^5 + m^3 \equiv 0 \mod 83 \]
or equivalently,
\[ m^{3^{4n}+6} - m^{3^{4n}+4} \equiv m^5 - m^3 \mod 83 \]
\[ m^{3^{4n}+4}(m^2 - 1) \equiv m^3(m^2 - 1) \mod 83 \]
\[ m^{3^{4n}+4} \equiv m^3 \mod 83 \]
\[ m^{3^{4n}+1} \equiv 1 \mod 83 \]

Since 83 is prime, there exists a primitive root \( g \) in \( \mathbb{Z}_{83} \). Let \( m = g \):
\[ g^{3^{4n}+1} \equiv 1 \mod 83 \]

Since \( g \) is a primitive root, the order of \( g \) is 82. Therefore, we need:
\[ 3^{4n} + 1 \equiv 0 \mod 82 \]
\[ 3^{4n} \equiv -1 \mod 82 \]

Since \( 3^{41} \equiv -1 \mod 82 \), we have:
\[ 3^{4n} \equiv -1 \mod 82 \]
\[ (-1)^n \equiv -1 \mod 82 \]

Thus, \( n \) must be odd. This condition is also sufficient because if \( n \) is odd, then \( 3^{4n} \equiv -1 \mod 82 \) and \( f(m, n) \equiv 0 \mod 83 \).

Therefore, the answer is every odd \( n \).

The final answer is every odd \( \boxed{ n } \).