Does there exist a natural number $ n$ such that $ n>2$ and the sum of squares of
some $ n$ consecutive integers is a perfect square?
To determine if there exists a natural number \( n \) such that \( n > 2 \) and the sum of squares of some \( n \) consecutive integers is a perfect square, we need to analyze the sum of squares of \( n \) consecutive integers.

1. Let the \( n \) consecutive integers be \( m+1, m+2, \ldots, m+n \). The sum of these integers is:
   \[
   S = (m+1) + (m+2) + \cdots + (m+n)
   \]

2. Using the formula for the sum of an arithmetic series, we have:
   \[
   S = \frac{(m+1 + m+n)}{2} \cdot n = \frac{(2m+n+1)}{2} \cdot n
   \]

3. Simplifying the expression, we get:
   \[
   S = \frac{n(2m+n+1)}{2}
   \]

4. Now, we need to check if this sum \( S \) can be a perfect square. Let's consider the case when \( n = 2k+1 \) (an odd number):
   \[
   S = (2k+1)(k+m+1)
   \]

5. If we take \( m = k \), then:
   \[
   S = (2k+1)(k+k+1) = (2k+1)(2k+1) = (2k+1)^2
   \]

6. This shows that for every \( k \), there are \( 2k+1 \) consecutive integers such that their sum is a perfect square.

7. For example, if \( k = 2 \), then \( n = 5 \):
   \[
   3 + 4 + 5 + 6 + 7 = 25 = 5^2
   \]

Thus, we have shown that there exists a natural number \( n \) such that \( n > 2 \) and the sum of squares of some \( n \) consecutive integers is a perfect square.

The final answer is \( \boxed{ n = 2k+1 } \) for any natural number \( k \).