$2AB$ is a perfect square and $A+B$ is not a perfect square If :$A=1!2!\cdots 1002!$, and $B=1004 ! 1005!\cdots2006!$, how to prove that:
a) $2AB$  is a perfect square 
b) $A+B$ is not a perfect square 
 A: Let $e(n,p)$ denote the exponent of a prime $p$ in the number $n$.
It is well-known that
$$e(n!,p):=\left\lfloor\frac np\right\rfloor+\left\lfloor\frac n{p^2}\right\rfloor+\left\lfloor\frac n{p^3}\right\rfloor+\ldots.$$
For part b), let $p$ be a prime with $2p\le 1002<3p$ (and of course $p^2>1002$). 
This is equivalent to $334<p\le501$ and a prime in this range is readily found, e.g. $p=337$ or $p=499$.
Then $e(n!,p)=0$ for $n<p$, $e(n!,p)=1$ for $p\le n<2p$, $e(n!,p)=2$ for $2p\le n\le 1002$. Therefore the exponent of $p$ in $A$ is given by $$e(A,p):=\sum_{n=1}^{1002} e(n,p) = p\cdot1+(1003-2p)\cdot 2=2006-3p.$$ Clearly $$e(B,p):=\sum_{n=1004}^{2006} e(n,p)\ge(2006-1003)\cdot2=2006>e(A,p).$$ Therefore, we have $e(A+B,p)=\min\{e(A,p), e(B,p)\}=e(A,p)$ is odd, hence $A+B$ cannot be a square. 
A: Hint for a
$$x!(x+1)!=[x!]^2 (x+1)$$
It follows that
$$A= (..)^2 \cdot 2 \cdot 4 ... \cdot 1002= (...)^2 \cdot 2^{501} \cdot 501!  \,.$$
do the same to $B$, which has an odd number of terms and you are done.
