<s>[INST] Hard problem I can not prove properly. I managed to concoct a weak proof for the below problem but I would love to see if there are any different ways to solve it.
Find the largest perfect square that divides $2014^3-2013^3+2012^3-2011^3 .... +2^3-1^3$
the options listed are 
a)$1^2$
b)$2^2$
c)$1007^2$
d)$2014^2$
I think the answer is c) but i just got that by calculating the last digit of the sum to check the divisibility. But I doubt that the proof is concrete. Help would be appreciated.
 [/INST] Consider:
\begin{align}
A_{2014} &= \sum_{n=1}^{1007} [(2n)^3 - (2n-1)^3] = \sum_{n=1}^{1007} ((2n) -(2n-1))((2n)^2 + (2n)(2n-1) + (2n-1)^2) \\
&= \sum_{n=1}^{1007} (12n^2 - 6n + 1) \\
&= 4087631519 = 19^{2} \cdot 29^{1} \cdot 53^{2} \cdot 139 = 29 \cdot 139 \cdot (1007)^2.
\end{align}
In comparison: $2014^{2} = 2^2 \cdot 19^2 \cdot 53^2$, $1007^{2} = 19^2 \cdot 53^2$
</s>