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There is series like $N = 3! + 4! +\cdots+ 64!$. It is asked whether it is perfect square or cube . How to identify Whether $N$ is perfect square or cube for any big factorial or for sum of factorial ?

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    $\begingroup$ It's not divisible by $4$. $\endgroup$ Commented Oct 19, 2018 at 19:34
  • $\begingroup$ @LordSharktheUnknown ...but it is divisible by $2$. Slick. $\endgroup$ Commented Oct 19, 2018 at 19:37

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Let us define $a_n$ as $$a_n=3!+4!+...+n!$$ Rearrange this sum as $$a_n=3!(1+4+4\cdot 5+4\cdot 5\cdot 6+...+4\cdot5\cdot ...\cdot n)$$ Since $3!=6$ is not a perfect square or divisible by a perfect square, in order for $a_n$ to be a perfect square, the sum $1+4+4\cdot 5+...+4\cdot5\cdot ...\cdot n$ must be divisible by $6$. Note that all terms including and after the $4\cdot 5\cdot 6$ term are divisible by $6$, so the whole sum is divisible by $6$ if and only if $1+4+4\cdot 5$ is divisible by $6$. It is not divisible by $6$, so $a_n$ cannot be a perfect square.

Of course, this makes the assumption that $n\ge 6$, but you can check the cases $n=3,4,5$ for yourself since they are only finitely many.

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the number is of form 8k+6. a perfect square cant be of this form.

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  • $\begingroup$ This looks like a perfectly OK answer to me. It does not give all the details; so what? It is a very strong hint, like many other answers on this site. $\endgroup$
    – David K
    Commented Oct 19, 2018 at 20:31

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