All you need is a 4d imagination

1.- $a = b = 0 \Rightarrow a^2 + kab + b^2 =0 \space, \forall k \in \mathbb{R}$ . This is trivial.

2.- Now, consider $a = \text{C = Constant}$ , and we are going to solve the equation $A \equiv b^2 + kab + a^2 = 0$ with $b$ unknown. $b^2 + kab + a^2 = 0 \iff b = \frac{ -ka \pm \sqrt{k^2a^2 - 4a^2}}{2}$ Then, paying attention to the discriminant $k^2a^2 - 4a^2$ then we have 3 possibilities:

2.a) $k = \pm 2 \Rightarrow (b - a)^{2} = 0, \space \vee \space, (b + a)^{2} = 0 \Rightarrow b = \pm a$ is solution (the only solution (one solution for each value of $k$ )) of $A$ but it doesn't imply $a = b = 0$

2.b) $|k| > 2 \Rightarrow$ the discriminant of $A$ is bigger than $0$ and we have two real solutions for $A$ and these solutions doesn't lead to $a = b = 0$

2.c) $|k| < 2$ . In this case, the discriminant of $A$ is smaller than $0$ and this implies that the only solutions for $A$ are imaginary solutions, or in other words, the equation (or equality in this case) solely is got iif $a = b = 0$ . Therefore, $(A \equiv a^2 + kab + b^2 = 0 \Rightarrow a = b = 0) \iff k \in (-2, 2)$

Any linear combination of $a^2 + b^2$ and $ab$ can be written as a linear combination of $(a+b)^2$ and $(a-b)^2$ , and it is not difficult to see that $a^2 + kab + b^2 = \frac14\left[(k+2)(a+b)^2 - (k-2)(a-b)^2\right].$ In order for this to be zero, we must have both terms equal, and therefore certainly of equal sign.

• Both terms zero: either $a+b = a-b = 0$ (the trivial solution), or $k = \pm 2$ and $a \pm b = 0$ . The latter has non-trivial solutions, namely all pairs $(x, \pm x)$ .

• $k+2$ and $k-2$ have the same sign: it is easy to see that this is the case if $k < -2$ or $k > 2$ . The values of $a+b$ and $a-b$ can always be tweaked to make both terms equal.

Thus we there are non-trivial solutions if $k \leq -2$ or $k \geq 2$ ; the desired interval is $(m,n) = (-2,2)$ ; and we submit the answer as $(-2)^2 + 2^2 = \boxed{8}$ .