Find all functions $f,g : N \to N$ such that for all $m ,n \in  N$ the following relation holds: $$f(m ) - f(n) = (m - n)(g(m) + g(n))$$. 
Note: $N = \{0,1,2,...\}$
To find all functions \( f, g : \mathbb{N} \to \mathbb{N} \) such that for all \( m, n \in \mathbb{N} \), the following relation holds:
\[ f(m) - f(n) = (m - n)(g(m) + g(n)), \]
we will proceed step-by-step.

1. **Substitute \( n = 0 \) into the given equation:**
   \[ f(m) - f(0) = m(g(m) + g(0)). \]
   This can be rewritten as:
   \[ f(m) = f(0) + m(g(m) + g(0)). \]

2. **Substitute \( n = 1 \) into the given equation:**
   \[ f(m) - f(1) = (m - 1)(g(m) + g(1)). \]
   Using the expression for \( f(m) \) from step 1, we get:
   \[ f(0) + m(g(m) + g(0)) - f(1) = (m - 1)(g(m) + g(1)). \]

3. **Simplify the equation from step 2:**
   \[ f(0) + m(g(m) + g(0)) - f(1) = mg(m) + mg(1) - g(m) - g(1). \]
   Rearrange terms to isolate \( g(m) \):
   \[ f(0) - f(1) + mg(0) + mg(1) - g(1) = mg(m) - g(m). \]
   \[ f(0) - f(1) + mg(0) + mg(1) - g(1) = (m - 1)g(m). \]

4. **Assume \( g(m) \) is a linear function:**
   Let \( g(m) = am + b \), where \( a \) and \( b \) are constants. Substitute this into the equation:
   \[ f(0) - f(1) + m(g(0) + g(1)) - g(1) = (m - 1)(am + b). \]
   Simplify:
   \[ f(0) - f(1) + m(g(0) + g(1)) - g(1) = am^2 + bm - am - b. \]

5. **Determine the form of \( f(m) \):**
   From step 1, we have:
   \[ f(m) = f(0) + m(g(m) + g(0)). \]
   Substitute \( g(m) = am + b \):
   \[ f(m) = f(0) + m(am + b + g(0)). \]
   \[ f(m) = f(0) + am^2 + m(b + g(0)). \]
   Let \( c = f(0) \) and \( d = b + g(0) \):
   \[ f(m) = am^2 + dm + c. \]

6. **Verify the solution:**
   Substitute \( f(m) = am^2 + dm + c \) and \( g(m) = am + b \) back into the original equation:
   \[ f(m) - f(n) = (am^2 + dm + c) - (an^2 + dn + c) = a(m^2 - n^2) + d(m - n). \]
   \[ (m - n)(g(m) + g(n)) = (m - n)(am + b + an + b) = (m - n)(a(m + n) + 2b). \]
   Both sides are equal, confirming the solution.

The final answer is \( \boxed{ (f(n), g(n)) = (an^2 + 2bn + c, an + b) } \).