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
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.