The sum of n+1 consecutive squares, beginning with the square of

n(2n+1),
is equal to the sum of the squares of the next n consecutive integers.

**
n=1 3**^{2} + 4^{2} = 5^{2}
n=2 10^{2} + 11^{2} + 12^{2} = 13^{2} + 14^{2}
.
.
.

Prove that the proposition holds for all integers greater than zero.

By the problem, we have:

y^2 + Sum (i= 1 to n) (y+i)^2

= Sum (i= n+1 to 2n) (y+ 2i)^2, for n, y > 0

Substituting y = x-n, we have:

x^2 + Sum (i= 1 to n) (x-i)^2 = Sum (i= 1 to n) (x+i)^2

-> x^2 = Sum (i= 1 to n) (x+i)^2 - Sum (i= 1 to n) (x-i)^2

-> x^2 = 4x(1+2+….n)

-> x^2 = 2xn(n+1)

Thus, x = 0, or : x= 2n(n+1)

If x=0, then y = -n < 0, which is a contradiction.

Thus, x = 2n(n+1), giving: y = 2n(n+1) – n = n(2n+1).

Hence the proof.

For the sufficiency part, it can be subsequently verified that :

(n(2n+1))^2 + Sum (i= 1 to n) ((n(2n+1))+i)^2

= Sum (i= n+1 to 2n) ((n(2n+1))+ 2i)^2, whenever n is an integer > 0

*Edited on ***September 27, 2008, 6:37 am**