Find all conditions for $x$ that the equation $1\pm 2 \pm 3 \pm 4 \pm \dots \pm n=x$ has a solution. Find all conditions for $x$ so that the equation $1\pm 2 \pm 3 \pm 4 \pm \dots \pm 1395=x$ has a solution.
My attempt: $x$ cannot be odd because the left hand side is always even then we have $x=2k(k \in \mathbb{N})$ also It has a maximum and minimum 
$1-2-3-4-\dots-1395\le x \le 1+2+3+4+\dots +1395$
But I can't show if these conditons are enough or make some other conditions.

The solution for $n$ as in the title is as follows:
Let $p = (n-3)(n+4)/2$ then the solutions for x are  $-p-4, -p, -p+2, \ldots,p-2, p, p +2, p+4$  where from the term $-p$ on we go to the term $p$ each time adding $2$. So only the first and the last two terms are irregular. We can verify:


*

*for $n = 4$ we have $-8, -4, -2, 0, 2, 4, 6, 10$ with p = $(4-3)(4+4/2 = 4)$

*for $n = 5$ we have $ -13, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 15$ with $p = (5-3)(5+4)/2 = 9$

*for $n = 1395$ we will have $-973708, -973704, -973702, \text{ each time add 2 , 973704, 973706, 973710}$ with $p = (1935-3)(1935+4)/2 = 973704$



Proof: we have to show that, having established the series for $n$ then we obtain the new series by adding and subtracting $n+1$ to all previous numbers. Indeed we have that $p' = (n'-3(n'-4)/2 = p + n + 1$ for $n' = n+1$.
