Set of $n$ integers with all subsets a perfect power? I am new to number theory, and I am stuck on this question:
Prove that for all integers $n$, there exists a set of $n$ integers, call it $S$, such that the sum of any subset of $S$ is a perfect power. Describe this set for $n=2017$.
I think that if I write it in the form:
$\{x,2x,3x,\cdots, nx\}$ where for $1 < k < (1 + 2 + 3 + \cdots+ n)$ $kx$ is a perfect power, I should be able to describe $x$ (and find it for $n=2017$), but I'm not sure how. 
I feel like I should be able to use the Chinese remainder theorem in some way, but I'm not sure how to apply the fact that values need to be perfect powers while thinking about factors remainders.
Please help!
 A: Let $S = \{1x, 2x, \dots , 2017x\}$ where $x$ is a a positive integer number to be determined.
Let $s_1,s_2,\dots, s_m$ be the sum of any non-empty subset of $\{1, 2, \dots , 2017\}$. It suffices to show that there is $x$ such that $x\cdot s_i$ is a perfect power for all $i=1,\dots,m$.
Let $p_1^{a_{1i}}\cdots p_{r}^{a_{ri}}$ be the prime factorization of $s_i$ (here $p_i$ denotes the $i$th prime) and let $p_1^{x_{1}}\cdots p_{r}^{x_{r}}$ be the prime factorization of $x$. Then $x\cdot s_i$ is a perfect power iff
$$g_i:=\gcd(a_{1i}+x_1,\dots,a_{ri}+x_r)\geq 2.$$
Now for each $j=1,\dots,r $, the system of congruences 
$$\begin{cases}
x_j\equiv -a_{j1} \pmod{p_1}\\
x_j\equiv -a_{j2} \pmod{p_2}\\
\dots\\
x_j\equiv -a_{jm} \pmod{p_m}\\
\end{cases}$$
has a solution by the Chinese Remainder Theorem. By taking $x_j$ in this way, it follows that $p_i$ divides $g_i$ which implies that $g_i\geq 2$ and we are done.
P.S. Note that asking for a set $S$ such that the sum of any subset of S is NOT a perfect power is much easier.
In fact, let $p$ be a prime with $p > \sum_{k=1}^{2017}k=2017\cdot 1009$
and let 
$$S = \{p, 2p, 3p, \dots , 2017p\}.$$
Then the sums of any non-empty subsets of $S$ is equal to $pm$, with 
$$m \leq \sum_{k=1}^{2017}k=2017\cdot 1009.$$
which implies that $m$ is not divisible by $p$. So we may conclude that the sum $mp$ is not a perfect power.
