2. Find all natural $n>1$ for which value of the sum $2^2+3^2+...+n^2$ equals $p^k$ where p 
is prime and k is natural
To find all natural numbers \( n > 1 \) for which the value of the sum \( 2^2 + 3^2 + \ldots + n^2 \) equals \( p^k \) where \( p \) is a prime and \( k \) is a natural number, we start by using the formula for the sum of squares of the first \( n \) natural numbers:

\[
\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}
\]

We need to find the sum from \( 2^2 \) to \( n^2 \):

\[
\sum_{k=2}^{n} k^2 = \sum_{k=1}^{n} k^2 - 1^2 = \frac{n(n+1)(2n+1)}{6} - 1 = \frac{n(n+1)(2n+1) - 6}{6}
\]

We set this equal to \( p^k \):

\[
p^k = \frac{n(n+1)(2n+1) - 6}{6}
\]

We can rewrite the numerator:

\[
n(n+1)(2n+1) - 6 = 2n^3 + 3n^2 + n - 6 = (n-1)(2n^2 + 5n + 6)
\]

Thus, we have:

\[
p^k = \frac{(n-1)(2n^2 + 5n + 6)}{6}
\]

We now consider two cases:

**Case 1: \( p \nmid n-1 \)**

In this case, \( n-1 \) must be one of the divisors of 6 (since the denominator is 6 and we need the result to be an integer). The possible values for \( n-1 \) are 1, 2, 3, and 6. Therefore, the possible values for \( n \) are:

\[
n = 2, 3, 4, 7
\]

We check each value:

- For \( n = 2 \):

\[
\sum_{k=2}^{2} k^2 = 2^2 = 4 = 2^2
\]

- For \( n = 3 \):

\[
\sum_{k=2}^{3} k^2 = 2^2 + 3^2 = 4 + 9 = 13
\]

13 is not a power of a prime.

- For \( n = 4 \):

\[
\sum_{k=2}^{4} k^2 = 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29
\]

29 is not a power of a prime.

- For \( n = 7 \):

\[
\sum_{k=2}^{7} k^2 = 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 = 4 + 9 + 16 + 25 + 36 + 49 = 139
\]

139 is not a power of a prime.

Thus, the only valid solution in this case is \( n = 2 \).

**Case 2: \( p \mid n-1 \)**

Let \( n-1 = px \) for some integer \( x \). We rewrite the expression in terms of \( px \):

\[
p^k = \frac{(n-1)(2n^2 + 5n + 6)}{6} = \frac{px(2(px)^2 + 5px + 13)}{6}
\]

For \( p^k \) to be an integer, \( 2(px)^2 + 5px + 13 \) must be divisible by 6. However, since \( p \) is a prime, we need \( p \mid 13 \), which implies \( p = 13 \). Testing this does not yield any valid solutions, as the resulting expression is too large to be a factor of 6.

Therefore, there are no solutions in this case.

The final answer is \( \boxed{ n = 2 } \).