Parallelego my eggo

For $f(x)$ to have four roots forming a parallelogram, there must exist a complex number $z$ (the centre of the parallelogram) such that $f(x+z)$ has distinct roots $\pm \alpha$ , $\pm \beta$ , and hence $f(x+z) \; = \; (x^2 - \alpha^2)(x^2-\beta^2) \; = \; x^4 - (\alpha^2+\beta^2)x^2 + \alpha^2\beta^2$ has no terms in $x^3$ or $x$ . To make the term in $x^3$ vanish, we must have $z=\tfrac32$ . The coefficient of $x$ in $f(x+\tfrac32)$ is $-6(a^2-4a+3) = -6(a-1)(a-3)$ , so that $a=1,3$ .

When $a=1$ , the roots of $f(x)$ are repeated, being $\tfrac12(3\pm\sqrt{5})$ , both twice. The roots of $f(x)$ are distinct (two real, two complex) when $a=3$ . Thus the required answer is $3^2=9$ .

Let $z_{1} , z_{2} , z_{3}$ , and $z_{4}$ be the roots of the equation represented on parallelogram ABCD in argand plane.

Clearly , midpoint of AC is same as midpoint of BD .

$\Rightarrow \frac{z_{1} + z_{3}}{2} = \frac{z_{2} + z_{4}}{2}$

Now rearranging vieta's formulae in a useful manner, $z_{1} + z_{2} + z_{3} + z_{4} = 6 \Rightarrow \boxed{z_{1} + z_{3} = z_{2} + z_{4} = 3}$ .

Also , $z_{1}(z_{2} + z_{4}) + z_{3}(z_{2} + z_{4}) + z_{1}z_{3} + z_{2}z_{4} = 11a$

$\Rightarrow z_{1}z_{3} + z_{2}z_{4} = 11a - 9$

also , $z_{1}z_{3}(z_{2} + z_{4}) + z_{2}z_{4}(z_{1} + z_{3}) = 3(2a^2 + 3a - 3 )$

Using $z_{1} + z_{3} = z_{2} + z_{4} = 3,$ and $z_{1}z_{3} + z_{2}z_{4} = 11a - 9$ ,

$2a^2 + 3a - 3 = 11a - 9$ or $a^2 - 4a + 3 = 0$ $\Rightarrow a=1,3$

But for $a=1$ , the roots won't be distinct , hence, $a$ can be $3$ only , hence $a^2 = \fbox{9}$

The only way to form a parallelogram with the solutions is if we have solutions of the form $x=p\pm iq$ and $x=p\pm r$ where $p, q$ and $r$ are real numbers. The function must therefore be of the form: $f_a(x)=(x-(p+iq))(x-(p-iq))(x-(p+r))(x-(p-r))$ $\qquad=x^4-4px^3+(6p^2+q^2-r^2)x^2-(4p^3+2p(q^2-r^2))x+p^4+p^2(q^2-r^2)-q^2r^2$ If we now compare the coefficients of each power of $x$ with the equation given in the question then we get: \begin{aligned} 4p&=6\tag{1}\\ 6p^2+q^2-r^2&=11a\tag{2}\\ 4p^3+2p(q^2-r^2)&=3(2a^2+3a-3)\tag{3}\\ p^4+p^2(q^2-r^2)-q^2r^2&=1\tag{4} \end{aligned} Equation (1), (2) and (3) can be used to deduce: $4a^2-16a+12=0\implies (a-1)(a-3)=0\implies a=1\text{ or }a=3$ Equation (4) can then be used to show that $a=1$ leads to $q^2r^2=-\frac{25}{16}$ and $a=3$ leads to $q^2r^2=\frac{767}{16}$ .

But $q$ and $r$ are real numbers, hence $a=3$ is the only viable solution, hence sum of squares of $a=3^2=9$ .