Let $n,a_1,a_2,\ldots,a_n$ be integers such that $a_1a_2a_3\cdots a_n =n$ and $a_1+a_2+a_3+\cdots+a_n=0$. Prove that 4 divides n

Question

I need help with this problem: Let $n,a_1,a_2,\ldots,a_n$ be integers such that $a_1a_2a_3\cdots a_n =n$ and $a_1+a_2+a_3+\cdots+a_n=0$. Prove that 4 divides n

I know that $a_i$ divides $n$ for any $i=1,2,3,\ldots,n$ and that $n>0$. So, there must be pair number of negative integers, but I don't know how to start the proof. I'm new at number theory. Need help please

Answer 1

If $n$ is odd, then each of the $a_i$ are odd. Then $a_1+a_2+\cdots+a_n$ is the sum of an odd number of odd numbers, which cannot equal zero (since zero is even)! So $n$ can't be odd.

Now if $n$ is a multiple of $2$ but not $4$, exactly one of the $a_i$ must be even and all the others must be odd. (If there were two or more even factors then $a_1a_2\cdots a_n$ would be a multiple of 4.) That means $a_1+a_2+\cdots+a_n$ is the sum of one even number and $n-1$ odd numbers, but since $n-1$ itself is odd, the total sum must be odd and thus cannot be zero again!

This leaves only the possibility that $n$ is a multiple of 4, and we are done.