Show that if you sum 9999 consecutive squares, the result cannot be a perfect power.

Let the squares to be summed be denoted by be (y + i)^2, where:

i = -4999, -4998, -4997, ......., -3, -2, -1, 0, 1,

2, 3, ......., 4998, 4999

Then, by conditions of the problem:

y^2 + Sum (j = 1 to 4999) [(y+j)^2 + (y-j)^2] = z^p, where z is an

integer with p> =2

Or, 9999*(y^2) + (4999*5000*9999)/3 = z^p

Or, (3*11*101) [3*y^2 + (4999)*(5000)] = z^p.

This is possible only when z=3a for some integer a.

Accordingly, we have:

(11*101) [3*y^2 + (4999)*(5000)] = 3^(p-1)*(a^p).....(*)

Since p> =2, it follows that the rhs is divisible by 3.

However, neither 4999 nor 5000 is divisible by 3. Also:

(3, 11) = (3, 101) = 1

Thus, the lhs of the relationship (*) is NOT divisible by 3.

This is a contradiction.

Consequently, the sum of sum 9999 consecutive squares can NEVER correspond to a perfect power.