4 Find a positive integer $N$ and $a_{1}, a_{2}, \ldots, a_{N}$, where $a_{k}=1$ or $a_{k}=-1$ for each $k=1,2, \ldots, N$, such that
$$
a_{1} \cdot 1^{3}+a_{2} \cdot 2^{3}+a_{3} \cdot 3^{3}+\cdots+a_{N} \cdot N^{3}=20162016,
$$
or show that this is impossible.
Solution: It is possible, as long as the sum $S$ desired is a multiple of 48 , with $N=S / 6$, which in this case is 3360336 , and the $a_{k}$ repeats the 8 -term pattern $-1,1,1,-1,1,-1,-1,1$.
Use the observation that if $f(x)$ is a degree- $k$ polynomial, then for any constant $h$, the difference $f(x+h)-f(x)$ will be a degree- $(k-1)$ polynomial. If we iterate this process three times, we can find a way to manipulate consecutive cubes to always get a constant.
More precisely, let
$$
c_{0}, c_{1}, c_{2}, c_{3}, c_{4}, c_{5}, c_{6}, c_{7} \ldots
$$
be consecutive cubes. In other words, $c_{m}=(m+u)^{3}$ for some fixed starting integer $u$. Then the differences
$$
c_{1}-c_{0}, \quad c_{3}-c_{2}, \quad c_{5}-c_{4}, \quad c_{7}-c_{6}, \ldots
$$
will be quadratic functions; i.e., if we define $a_{m}:=c_{m+1}-c_{m}$, then $a_{m}$ is a quadratic function of $m$ (depending on the parameter $u$, as well), and the differences are $a_{0}, a_{2}, a_{4}, a_{6}, \ldots$. Continuing, we see that the differences
$$
a_{2}-a_{0}, \quad a_{6}-a_{4}, \ldots
$$
will be a linear sequence; i.e., $b_{m}:=a_{m+2}-a_{m}$ is a linear function of $m$ (with parameter $u$ ), and our differences are $b_{0}, b_{4}, \ldots$. Finally, the sequence
$$
b_{4}-b_{0}, \quad b_{12}-b_{8}, \ldots
$$
is constant, no matter what the parameter $u$ equals! We have
$$
b_{4}-b_{0}=a_{6}-a_{4}-\left(a_{2}-a_{0}\right)=a_{6}-a_{4}-a_{2}+a_{0}=c_{7}-c_{6}-c_{5}+c_{4}-c_{3}+c_{2}+c_{1}-c_{0},
$$
and since this is constant, we can compute it using any value of $u$. Taking $u=-3$, the constant must equal
$$
4^{3}-3^{3}-2^{2}+1^{3}-0^{3}+(-1)^{3}+(-2)^{3}-(-3)^{3}=48 \text {. }
$$

In other words, if we define
$$
s_{u}:=-u^{3}+(u+1)^{3}+(u+2)^{3}-(u+3)^{3}+(u+4)^{3}-(u+5)^{3}-(u+6)^{3}+(u+7)^{3},
$$
then $s_{u}=48$ for all values of $u$.
Since $20162016=420042 \cdot 48$, we can easily write 20162016 as a sum of 420042 8-element sum/differences of $420042 \cdot 8=3360336$ consecutive cubes:
$$
\begin{aligned}
20162016=\sum_{k=0}^{420041} s_{8 k+1}= & -1^{3}+2^{3}+3^{3}-4^{3}+5^{3}-6^{3}-7^{3}+8^{3} \\
& -9^{3}+10^{3}+11^{3}-12^{3}+13^{3}-14^{3}-14^{3}+15^{3} \\
& \cdots \\
& -3360329^{3}+3360330^{3}+\cdots-3360335^{3}+3360336^{3} .
\end{aligned}
$$

NOTE: About half of the correct solutions were purely computational, making use of the fact that $1^{3}+2^{3}+\ldots+n^{3}=(n(n+1))^{2} / 4$ and putting $n=95$ gets you to a number that is rather close to 20162016. It is then possible to work out-with great difficulty, by hand-values that work. Clearly this method is not generalizable, unlike the multiple-of- 48 method above.