Answer: All multiples of $4$. First lets see why $4|n$ is necessary. Observet that $a_i\neq 0$ since $\prod a_i=n\neq 0$ Suppose $n$ is odd, then all $a_i$'s are odd and thus $\sum a_i=1\pmod 2$ which is impossible since $\sum a_i=0$. Now assume that $n=2t$ where $t$ is odd. Since $\prod a_i$ is even but not a multiple of $4$ there exist exactly one $j$ such that $a_j$ is even and $a_r$ is odd whenever $r\neq j$ But now since $2|n$ we have $\sum a_i=1\pmod 2$ a contradiction. It remains to show that $n=4k$ works. Suppose $n=2^ab$ where $a\geq 2$ and $b$ is odd. If $a$ is even take $a_1=2,a_2=-2,..,a_{a-1}=2,a_{a}=-2,a_{a+1}=b,a_{a+2}=-1,a_{a+3}=-1,..,a_{a+b-1}=-1$ and now $2^ab-a-b+1=even$ number of $a_i$'s remain, so just take alternately $+1,-1$ If $a$ is odd take $a_1=4,a_2=-2..,a_{a-1}=-2,a_{a}=2,a_{a+1}=b,a_{a+k}=-1,k=2,..,a+b+1$ and again an even number of indices remain so take again ternately $+1,-1$

smartvong wrote: It is known that there are $n$ integers $a_1, a_2, \cdots, a_n$ such that $$a_1 + a_2 + \cdots + a_n = 0 \qquad \text{and} \qquad a_1 \times a_2 \times \cdots \times a_n = n.$$Determine all possible values of $n$. Prod55 wrote: Answer: All multiples of $4$. First lets see why $4|n$ is necessary. Uhhh ? What about $n=6$ with $-2,-1,-1,-1,1,1,3$ ?

you are using 7 numbers

Prod55 wrote: you are using 7 numbers Mwaaaaahhh ! My bad ! Really sorry