math

\(x^2 + m x + n = 0,~(m,n)\in \mathbb{Z} \\ \text{the only possible value for x is }x = -3\\ x^2 + m x + n = (x+3)^2 = x^2 + 6x + 9\\ m = 6\)

Since   x  = -3 is the only value that makes the quadratic  = 0.....then this has a "double" root

So...the sum  of the roots  =  -3 + -3  =  -6

And the sum of the roots  in the form ax^2 + bx + c  =  -b/a 

So...in our case, the sum of the roots are  =  - m / 1  =   -m

So

-m   = -6

m  = 6