What is the number of elements in $\{1,2,...,2018\}$ such that it can be represented by $ x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1 \:\:? $ What is the number of elements in $\{1,2,...,2018\}$ such that it can be represented by
$$ x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1 \:\:? $$
$x \in \mathbb{Z}$

Attempt :
First, $x=1$ then the sum will be $7$, which is an element of the set in concern. Of course $x=0$ then the sum will be $1$, also in the set.
For $x=-1$, the sum will also be $1$. 
So far we have 2 elements of the set.
For $|x|>1$, we can view the summation as a finite geometric series :
$$ S = 1 + x + .... + x^{6} $$
$$ Sx = x + x^{2} + .... + x^{7} $$
subtract both of the above then we can have
$$ S(x) = \frac{1 - x^{7}}{1-x}, \:\:\: |x| > 1 $$
The key is to find $x*$ so that $S(x*)$ is the maximum just below $2018$.
I can do this by checking manually from $x=\pm 2, \pm 3, \pm 4 $. Without a calculator, this is quite tedious. Found out that $x = \pm 4$ do not satisfy the condition. 

So the answer is $6$ numbers, is this sufficient? (I was thinking that we may also have to check the monotonicity of $S(x)$)
What techniques are better than this one? Thanks.

For  brevity let $x^6 +x^5+ x^4 +x^3+ x^2+x+1=f(x).$... For $x\in \Bbb Z$ we have:
$$(Ia).\quad |x|\geq 4\implies f(x)=(x^6+x^4+x^2)(1+1/x)+1\geq$$ $$\geq (x^6+x^4+x^2)(1-1/|x|)+1>$$ $$>  x^6(1-1/|x|)\geq$$ $$\geq 4^6(1-1/2)=64^2/2=2048>2018.$$
$$(Ib).\quad |x|\leq 3\implies |f(x)|\leq |x|^6+|x|^5+|x|^4|+|x|^3+|x|^2+|x|+1\leq$$ $$\leq f(3)=(3^7-1)/2=(3^8/3-1)/2=$$ $$=(81^2/3-1)/2=1093<2018.$$
$(II). $ Obviously $x\geq 0\implies f(x)\geq 1.$  And $x<0\implies 1+1/x\geq 0\implies f(x)=(x^6+x^4+x^2)(1+1/x)+1\geq 1.$
$(III). $ So the solution set is $\{0,\pm 1,\pm 2,\pm 3\}.$
