Find the least positive integers $m,k$ such that 

a) There exist $2m+1$ consecutive natural numbers whose sum of cubes is also a cube.

b) There exist $2k+1$ consecutive natural numbers whose sum of squares is also a square.

The author is Vasile Suceveanu
### Part (a)
We need to find the least positive integer \( m \) such that there exist \( 2m+1 \) consecutive natural numbers whose sum of cubes is also a cube.

1. Consider the sequence of three consecutive natural numbers: \( 3, 4, 5 \).
2. Calculate the sum of their cubes:
   \[
   3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216
   \]
3. Notice that \( 216 = 6^3 \), which is a perfect cube.

Thus, for \( m = 1 \), we have \( 2m+1 = 3 \) consecutive natural numbers whose sum of cubes is a cube.

### Part (b)
We need to find the least positive integer \( k \) such that there exist \( 2k+1 \) consecutive natural numbers whose sum of squares is also a square.

1. Suppose \( k = 1 \):
   \[
   1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 \quad (\text{not a perfect square})
   \]
2. Suppose \( k = 2 \):
   \[
   2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 4 + 9 + 16 + 25 + 36 = 90 \quad (\text{not a perfect square})
   \]
3. Suppose \( k = 3 \):
   \[
   4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2 = 16 + 25 + 36 + 49 + 64 + 81 + 100 = 371 \quad (\text{not a perfect square})
   \]
4. Suppose \( k = 4 \):
   \[
   7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2 + 13^2 + 14^2 + 15^2 = 49 + 64 + 81 + 100 + 121 + 144 + 169 + 196 + 225 = 1149 \quad (\text{not a perfect square})
   \]
5. Suppose \( k = 5 \):
   \[
   18^2 + 19^2 + 20^2 + 21^2 + 22^2 + 23^2 + 24^2 + 25^2 + 26^2 + 27^2 + 28^2 = 324 + 361 + 400 + 441 + 484 + 529 + 576 + 625 + 676 + 729 + 784 = 5929
   \]
   Notice that \( 5929 = 77^2 \), which is a perfect square.

Thus, for \( k = 5 \), we have \( 2k+1 = 11 \) consecutive natural numbers whose sum of squares is a square.

The final answer is \( \boxed{ m = 1 } \) and \( k = 5 \).