Posed in regional mathematics Olympiad 1995 Call a positive integer $n$ good if there are $n$ integers, positive or negative, and not necessarily distinct, such that their sum and product are both equal to $n$ 
For example, $8$ is good since 
$$\begin{align} 
      8 = 4\times2\times1\times1&\times1\times1\times(-1)\times(-1) \\
                                &\textrm{and}\\
      8 = 4+2+1+1&+1+1+(-1)+(-1)
\end{align}$$
Show that integers of the form $4k+1$ ($k\ge0$) and $4l$ ($l\ge2$) are good
 A: Numbers of the form $4k+1$ are good: take the integers
$$4k+1,\underbrace{1,\dots,1}_{2k\,\text{times}},\underbrace{-1,\dots,-1}_{2k\,\text{times}}$$
The next claim is that numbers of the form $4l$ are good. We consider two cases: $l$ even and $l$ odd. If $l$ is even, we have $4l=8k$ and we take
$$2,4k,\underbrace{1,\dots,1}_{6k-2\,\text{times}},\underbrace{-1,\dots,-1}_{2k\,\text{times}}$$
If $l\ge 3$ is odd, we have $4l=4(2k+1)$ and we take
$$-2,4k+2,\underbrace{1,\dots,1}_{6k+3\,\text{times}},\underbrace{-1,\dots,-1}_{2k-1\,\text{times}}$$
A: As “Your Ad Here” wrote, if $n=4k+1$, take the following numbers:
$$4k+1,\underbrace{1,\dots,1}_{2k\,\text{times}},\underbrace{-1,\dots,-1}_{2k\,\text{times}}.$$
If $n=4k$, where $k>1$, write $n=2^j q$, where $q$ is odd and $j\ge2$, and proceed by induction on $j$.
If $j=2$, $n=4k$ with $k$ odd. Then take these numbers:
$$-2, 2k,\underbrace{1,\dots,1}_{3k\,\text{times}},\underbrace{-1,\dots,-1}_{k-2\,\text{times}}.$$
Their product is positive, because there are $1+3k-2$ (evenly many, when $k$ is odd) negative numbers, and its absolute value is $4k$, so the product is $4k$.
Their sum is $-2+2k+3k-(k-2)=4k$.
If $j>2$, $n=2N$, where $N=2k$ is “good.” Let $a_i$ be the $N$ numbers with sum and product equal to $N$. Now consider these $2N$ numbers:
$$2, a_1,\dots,a_N,\underbrace{1,\dots,1}_{\frac{N}{2}-2\,\text{times}},\underbrace{-1,\dots,-1}_{\frac{N}{2}\,\text{times}}.$$
Their product is $2\left(\prod{a_i}\right)(-1)^{\frac{n}{2}}=2N=n$, because $N$ is even. 
Their sum is $2+\left(\sum a_i\right)+\frac{N}{2}-2-\frac{N}{2}=2+N+N-2=2N=n$.
(Note that $\frac{N}{2}-2=\frac{2k}{2}-2$ is at least zero, because $k>1$ was given.)
