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

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