Cyclic equations

Let us consider the equations for $a_1$ , $a_2$ & $a_3$ .
$a_2+a_3+a_4+\ldots= a_1\rightarrow Eq.1$
$a_1+a_3+a_4+\ldots= a_2\rightarrow Eq.2$
$a_1+a_2+a_4+\ldots= a_3\rightarrow Eq.3$
$Eq.1$ & $Eq.2$ may be rearranged as:
$-a_1+a_2+a_3+a_4+\ldots= 0\rightarrow Eq.A$
&, $a_1-a_2+a_3+a_4+\ldots= 0\rightarrow Eq.B$
Subtracting $Eq.A$ $from Eq. B$ , we get,
$2a_1-2a_2=0$
$\Rightarrow a_1-a_2=0$
$\boxed {a_1=a_2}$
Now, rearranging $Eq.3$ ,
$a_1+a_2-a_3+a_4+\ldots= 0\rightarrow Eq.C$
Subtracting $Eq.A$ from $Eq.C$ , we get,
$2a_1-2a_3=0$
$\Rightarrow a_1-a_3=0$
$\boxed {a_1=a_3}$
On repeating the process with other variables, we can see that,
$\boxed {a_1=a_2=a_3=a_4=\ldots=a_n}\rightarrow Eq.4$
But, we also know that the value of any one variable is equal to the sum of all the other variables.
Combining the above condition with $Eq.4$ leaves us with only one possible value that can be assigned to the variables while satisfying both the conditions at hand, i.e $0$ .
So, $\boxed {a_1=a_2=a_3=a_4=\ldots=a_n=0}$
Hence,
$\boxed {a_1+a_2+a_3+a_4+\ldots+a_n= [0+0+0+\ldots n times)] = 0}$

$a_1+a_2$ could be anything you want when $n=2$ , so it should be added to the problem that $n\ge 3, n\in\mathbb N$ .

its a silly problem, one just has to restate the question in one's words....does it matter that a negative coefficient appears in a given set of equations from n=1 to n=n where n is index that all equal to 0. in other words Matrix with diagonal of -1, has at least one solution, ... the answer is yes, if you can get that matrix to look like idetety matrix, then there is at least one solution....to adition of all a's 0

All that is fine and dandy, except that if you follow the logic of the problem [i.e. a(1) + a(2) + ... + a(n) - a(i) = a(i)], and then add each of the right hand side equations while substituting each for its longer form, you end up with (n-1) = [a(1) + a(2) + ... + a(n)]/[a(1) + a(2) + ... + a(n)] = 1; therefore, n = 2 for all a(i) regardless of what the equations "look" like. Following this, any value for a(i) works, as long as that the problem is limited to only two terms--a(1) = a(2).

Any assumptions beyond this are actually not valid. While 0 is a possible soultion, so are all real numbers.