Quadratic complex roots

The equation holds true for $x^2+4x+5=1\implies x=-2$ and

$x^2-3x+7=0$

The later equation has two distinct roots whose product is $7$ (since $3^2-4\times 1\times 7=-19\neq 0$ )

So, in all, there are $n=3$ roots, whose GM is $\sqrt[3] {-2\times 7}=\sqrt[3] {-14}$

Hence $m=-14$ and $mn=\boxed {-42}$ .

If $a^b=1$ , then:

• $a=1$ , $b\in\mathbb{C}$ ,
• $a=-1$ , $b=2k\ \{k\in\mathbb{Z}\}$
• $a\in\mathbb{C}\setminus \{0\}$ , $b=0$

For the first possibility,

$x^2+4x+5=1\iff (x+2)^2=0\iff x=\boxed{-2}$ .

For the second possibility,

$x^2+4x+5=-1\iff x=-2\pm\sqrt{2}i$ , but then $x^2-3x+7$ is not an even integer (it is $15\pm 7\sqrt{2}i$ ), so there are no solutions here.

For the third possibility,

$x^2-3x+7=0\iff x=\dfrac{3\pm\sqrt{3^2-4\cdot 1\cdot 7}}{2}=\boxed{\frac{3\pm\sqrt{19}i}{2}}$ .

For completeness, we check the value of $x^2+4x+5$ for these same values of $x$ , which gives $\dfrac{17\pm7\sqrt{19}i}{2}$ which is nonzero in either case, so both solutions are valid.

$\therefore$ There are $n=3$ roots, with geometric mean $\sqrt[3]{-2\cdot \dfrac{3+\sqrt{19}i}{2}\cdot \dfrac{3-\sqrt{19}i}{2}}=\sqrt[3]{-14}\implies m=-14$ , so the answer is $3\cdot -14=\color{#20A900}{\boxed{-42}}$ .

Note: To avoid finding and multiplying the roots in the final step, we may also use Vieta's formula on the last quadratic, which has the product of roots $\Sigma\alpha\beta=\dfrac{7}{1}$ , so that the geometric mean is $\sqrt[3]{-2\cdot 7}$ , as before.

The given equation holds true when $x^2 - 3x + 7 = 0$ and $x^2 + 4x + 5 \ne 0$ . $x^2 - 3x + 7 = 0$ has two conjugate complex roots $\alpha$ and $\overline \alpha$ and by Vieta's formula , their product $\alpha \overline \alpha = 7$ .

The equation also holds true when $x^2 + 4x + 5 = 1$ or $x^2 + 4x + 4 = (x+2)^2 = 0$ , which has only one real root of $x = -2$ .

Therefore there are three roots or $n=3$ and the geometric mean of the three roots is $\sqrt[3]{(-2)\alpha \overline \alpha} = \sqrt[3]{-14}$ . Therefore $nm = 3(-14) = \boxed{-42}$ .