Let $\mathbb{F}_p$ be the set of integers modulo $p$. Call a function $f : \mathbb{F}_p^2 \to \mathbb{F}_p$ [i]quasiperiodic[/i] if there exist $a,b \in \mathbb{F}_p$, not both zero, so that $f(x + a, y + b) = f(x, y)$ for all $x,y \in \mathbb{F}_p$.
Find the number of functions $\mathbb{F}_p^2 \to \mathbb{F}_p$ that can be written as the sum of some number of quasiperiodic functions.
1. **Define the Vector Space of Functions:**
   Consider the vector space \( V \) of functions from \(\mathbb{F}_p^2 \to \mathbb{F}_p\). Each function \( f \) in this space can be represented as a polynomial in two variables \( x \) and \( y \) with coefficients in \(\mathbb{F}_p\).

2. **Characterize Quasiperiodic Functions:**
   A function \( f \) is quasiperiodic with period \((a, b)\) if \( f(x + a, y + b) = f(x, y) \) for all \( x, y \in \mathbb{F}_p \). This implies that \( f \) is invariant under translations by \((a, b)\).

3. **Basis Functions for Quasiperiodic Functions:**
   Define the basis functions \( e_{(x_0, y_0)}^{(a, b)}(x, y) \) as:
   \[
   e_{(x_0, y_0)}^{(a, b)}(x, y) = \sum_{i=0}^{p-1} 1_{x = x_0 + a \cdot i, y = y_0 + b \cdot i}
   \]
   These basis functions span the space of quasiperiodic functions with period \((a, b)\).

4. **Orthogonal Complement:**
   To find the dimension of the subspace spanned by \( e_{(x_0, y_0)}^{(a, b)} \), consider the orthogonal complement. The orthogonal complement consists of all functions \( f: \mathbb{F}_p^2 \to \mathbb{F}_p \) such that:
   \[
   \sum_{0 \le x, y \le p-1} f(x, y) e_{(x_0, y_0)}^{(a, b)}(x, y) = 0
   \]
   This translates to:
   \[
   \sum_{r=0}^{p-1} f(x_0 + a \cdot r, y_0 + b \cdot r) = 0
   \]
   for all \( x_0, y_0, a, b \).

5. **Polynomial Representation:**
   Any function \( f \) can be uniquely written as:
   \[
   f(x, y) = \sum_{i=0}^{p-1} \sum_{j=0}^{p-1} c_{i,j} x^i y^j
   \]
   Plugging this into the orthogonality condition, we get:
   \[
   \sum_{i=0}^{p-1} \sum_{j=0}^{p-1} c_{i,j} \sum_{r=0}^{p-1} (x_0 + a \cdot r)^i (y_0 + b \cdot r)^j = 0
   \]

6. **Expanding Using Binomial Theorem:**
   Expanding each term using the binomial theorem, we have:
   \[
   \sum_{i=0}^{p-1} \sum_{j=0}^{p-1} c_{i,j} \sum_{e_1=0}^{p-1} \sum_{e_2=0}^{p-1} \binom{i}{e_1} \binom{j}{e_2} x_0^{i-e_1} y_0^{j-e_2} a^{e_1} b^{e_2} \sum_{r=0}^{p-1} r^{e_1+e_2} = 0
   \]

7. **Properties of Primitive Roots:**
   By properties of primitive roots, the last sum is always zero unless \( p-1 \mid e_1 + e_2 \) and \( e_1 + e_2 = p - 1 \). Thus, we get:
   \[
   \sum_{i=0}^{p-1} \sum_{j=0}^{p-1} c_{i,j} \sum_{e_1=0}^{p-1} \sum_{e_2=0}^{p-1} \binom{i}{e_1} \binom{j}{e_2} x_0^{i-e_1} y_0^{j-e_2} a^{e_1} b^{e_2} 1_{\{p-1 \mid e_1 + e_2\}} = 0
   \]

8. **Coefficient Analysis:**
   Since this is a polynomial in \( x_0, y_0 \), for it to vanish for all \( x_0, y_0 \), each coefficient must be zero. The coefficient of \( x^{k_1} y^{k_2} \) is:
   \[
   \sum_{0 \le e_1, e_2 \le p-1, p-1 \mid e_1 + e_2} \binom{e_1+k_1}{k_1} \binom{e_2+k_2}{k_2} c_{e_1+k_1, e_2+k_2} a^{e_1} b^{e_2} = 0
   \]

9. **Free Variables:**
   For \( k_1, k_2 \neq 0 \), this gives \( c_{e_1+k_1, e_2+k_2} = 0 \) for any \( e_1, e_2 \) where \( p-1 \mid e_1 + e_2 \) and \( e_1 + e_2 \ge p-1 \). This implies \( c_{x, y} = 0 \) for all \( x + y > p-1 \). There are no linear constraints on \( x + y \le p-1 \).

10. **Dimension Calculation:**
    There are \(\frac{(p-1)p}{2}\) free variables. The orthogonal complement has dimension \(\frac{(p-1)p}{2}\), and the original vector space has dimension \(\frac{p(p+1)}{2}\).

The final answer is \( \boxed{ p^{(p^2 + p)/2} } \)