Is this quantity possible? I need help for my problem. 
A set $M$ of $2015$ positive integers has the following condition: if $x_1 < x_2 < ... < x_n$ are elements of $M$ with $1 \leq n \leq 2015$, then neither $\sum_{i=1}^{n} x_i$ nor $\prod_{i=1}^{n} x_i$ are squares.
Is there a set M which has this condition?
I suppose there is no such $M$ but I can't prove it. So I hope you can help me and thank you for your suggestions.
 A: It certainly is possible, although it may be difficult to actually write down. Here are some hints to a (probably non-optimal) solution:


*

*There is nothing special about the number $2015$. In fact there exist sets of size $n$ with these conditions for every $n$. Proof by induction seems the way to go.

*The second condition is easy to satisfy. If we take all the elements to be coprime and non-square then no products of elements can be square.

*For the inductive step, suppose we have a set $M$ of size $n$ which satisfies the conditions. Let $$\mathcal M = \left\{\sum_{i=1}^m x_i:1\le m \le n,\ x_1<\cdots<x_m\in M\right\}$$be the set of all sums of elements of $M$. $\mathcal M$ is a finite set, so it contains a maximal element $N$. To construct a set of size $n+1$ which satisfies the conditions, we need to find a non-square integer $k$ which is coprime to every element of $M$, for which $k+m$ is not square for every $m\in\mathcal M$.

*Gaps between square numbers can be made arbitrarily long. If we can find some $k\in\mathbb N$ which is non-square, coprime to every element of $M$ and translates $\mathcal M$ into a gap between two large squares, then we'll be done.

*Let $$k = 1+N^2\prod_{x\in M}x^2$$Then $k$ is certainly coprime to every element of $M$, and $$\left(N\prod_{x\in M}x\right)^2<k+m<\left(1+N\prod_{x\in M}x\right)^2$$ for every $m \in \mathcal M$.

