Determine the smallest natural number $a\geq 2$ for which there exists a prime number $p$ and a natural number $b\geq 2$ such that
\[\frac{a^p - a}{p}=b^2.\]
To determine the smallest natural number \( a \geq 2 \) for which there exists a prime number \( p \) and a natural number \( b \geq 2 \) such that
\[
\frac{a^p - a}{p} = b^2,
\]
we start by rewriting the given equation as:
\[
a(a^{p-1} - 1) = pb^2.
\]
Since \( a \) and \( a^{p-1} - 1 \) are coprime (i.e., \(\gcd(a, a^{p-1} - 1) = 1\)), we have two cases to consider:

### Case 1: \( a = px^2 \) and \( a^{p-1} - 1 = y^2 \) for some positive integers \( x \) and \( y \).

From Mihăilescu's Theorem (also known as Catalan's conjecture), which states that the only solution in the natural numbers of the equation \( x^a - y^b = 1 \) for \( a, b > 1 \) is \( 3^2 - 2^3 = 1 \), it follows that \( p = 2 \). Thus, we have:
\[
a = 2x^2 \quad \text{and} \quad a - 1 = y^2.
\]
This leads to the equation:
\[
2x^2 - y^2 = 1. \tag{1}
\]
The minimal solution for equation (1) is \( (x, y) = (1, 1) \). However, this gives:
\[
a = 2 \quad \text{and} \quad b^2 = \frac{2^2 - 2}{2} = 1,
\]
which implies \( b = 1 \), contradicting \( b \geq 2 \).

The next minimal solution for equation (1) with \( x > 1 \) is \( (x, y) = (5, 7) \). Thus:
\[
a = 2 \cdot 5^2 = 50 \quad \text{and} \quad b^2 = \frac{50^2 - 50}{2} = 35^2,
\]
which means \( b = 35 \). Therefore, the least value of \( a \) in this case is \( 50 \).

### Case 2: \( a = x^2 \) and \( a^{p-1} - 1 = py^2 \) for some positive integers \( x \) and \( y \).

Consider \( (x, y, p) = (3, 2, 2) \):
\[
a = 3^2 = 9 \quad \text{and} \quad b = 6.
\]
This satisfies the condition since:
\[
\frac{9^2 - 9}{2} = \frac{81 - 9}{2} = 36 = 6^2.
\]
Thus, \( a = 9 \) works.

If \( x = 2 \), then \( a = 4 \) and:
\[
4^{p-1} - 1 = (2^{p-1} - 1)(2^{p-1} + 1) = py^2.
\]
Since \( p \mid 2^{p-1} - 1 \) (by Fermat's Little Theorem), we have:
\[
2^{p-1} - 1 = pr^2 \quad \text{and} \quad 2^{p-1} + 1 = s^2
\]
for some positive integers \( r \) and \( s \). Mihăilescu's Theorem implies \( p \in \{2, 4\} \), but since \( p \) is prime, \( p = 2 \). Thus:
\[
2y^2 = 4^{2-1} - 1 = 3,
\]
which is a contradiction.

Therefore, the least value of \( a \) in this case is \( 9 \).

Since \( 9 < 50 \), the smallest natural number \( a \geq 2 \) that satisfies the given condition is \( a = 9 \).

The final answer is \(\boxed{9}\).