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
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
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 and a prime in this range is readily found, e.g. $p=337$ or $p=499$. Then $e(n!,p)=0$ for $n , $e(n!,p)=1$ 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.
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.