On the Cartesian plane consider the set $V$ of all vectors with integer coordinates. Determine all functions $f : V \rightarrow \mathbb{R}$ satisfying the conditions:
(i) $f(v) = 1$ for each of the four vectors $v \in V$ of unit length.
(ii) $f(v+w) = f(v)+f(w)$ for every two perpendicular vectors $v, w \in V$
(Zero vector is considered to be perpendicular to every vector).
1. **Base Case Verification:**
   We start by verifying the base case for \( n = 1 \). According to condition (i), for the four unit vectors \([1,0]\), \([0,1]\), \([-1,0]\), and \([0,-1]\), we have:
   \[
   f([1,0]) = f([0,1]) = f([-1,0]) = f([0,-1]) = 1
   \]
   This satisfies the base case for \( n = 1 \).

2. **Inductive Hypothesis:**
   Assume that for all \( k \leq n \), the function \( f \) satisfies:
   \[
   f([k,0]) = f([0,k]) = f([-k,0]) = f([0,-k]) = k^2
   \]

3. **Inductive Step:**
   We need to show that the property holds for \( k = n+1 \). Consider the vector \([n,1]\). Using condition (ii) and the inductive hypothesis:
   \[
   f([n,1]) = f([n,0]) + f([0,1]) = n^2 + 1
   \]
   Similarly, for the vector \([1,-n]\):
   \[
   f([1,-n]) = f([1,0]) + f([0,-n]) = 1 + n^2
   \]
   Therefore:
   \[
   f([n,1]) = f([1,-n]) = n^2 + 1
   \]

4. **Combining Results:**
   Using the results from the previous step, we can find \( f([n+1,0]) \). Consider the vector \([n+1,1-n]\):
   \[
   f([n+1,1-n]) = f([n,1]) + f([1,-n]) = (n^2 + 1) + (n^2 + 1) = 2n^2 + 2
   \]
   Also, using the property of perpendicular vectors:
   \[
   f([n+1,1-n]) = f([n+1,0]) + f([0,-(n-1)])
   \]
   Since \( f([0,-(n-1)]) = (n-1)^2 \), we have:
   \[
   2n^2 + 2 = f([n+1,0]) + (n-1)^2
   \]
   Solving for \( f([n+1,0]) \):
   \[
   2n^2 + 2 = f([n+1,0]) + (n^2 - 2n + 1)
   \]
   \[
   f([n+1,0]) = 2n^2 + 2 - (n^2 - 2n + 1) = n^2 + 2n + 1 = (n+1)^2
   \]

5. **Conclusion:**
   By induction, we have shown that for all \( n \in \mathbb{N} \):
   \[
   f([n,0]) = f([0,n]) = f([-n,0]) = f([0,-n]) = n^2
   \]

6. **General Case:**
   For any vector \( u = [m,n] \) where \( m, n \) are integers, we use the additivity property:
   \[
   f([m,n]) = f([m,0]) + f([0,n]) = m^2 + n^2
   \]

Thus, the only function \( f \) that satisfies the given conditions is:
\[
f([m,n]) = m^2 + n^2
\]

The final answer is \( \boxed{ f([m,n]) = m^2 + n^2 } \)