Wolfram Alpha=Cheating

Dividing the equation throughout by $z^{4}$ yields

$2(z^{4}+\frac{1}{z^{4}})-3(z^{3}-\frac{1}{z^{3}})-12(z^{2}+\frac{1}{z^{2}})+12(z-\frac{1}{z})+22=0$

Now, let's try to write express everything in terms of $(z-\frac{1}{z})$ . Rewrite the equation above as:

$2[(z-\frac{1}{z})^{4}+4(z-\frac{1}{z})^{2}+2]-3[(z-\frac{1}{z})((z-\frac{1}{z})^{2}+3)]-12[(z-\frac{1}{z})^{2}+2]+12(z-\frac{1}{z})+22=0$

Now, we make the obvious substitution $a=z-\frac{1}{z}$ and rewrite the equation above as

$2(a^{4}+4a^{2}+2)-3a(a^{2}+3)-12(a^{2}+2)+12a+22=0$ .

Upon simplifying, we get

$2a^{4}-3a^{3}-4a^{2}+3a+2=0$ , which factorizes beautifully as

$(a-1)(a+1)(a-2)(2a+1)=0$ .

Solving $z-\frac{1}{z}=1,-1,2,-\frac{1}{2}$ yields $8$ possible solutions for $z$ , which are $\frac{1 \pm \sqrt{5}}{2}, \frac{-1 \pm \sqrt{5}}{2}, 1 \pm \sqrt{2}$ and $\frac{-1 \pm \sqrt{17}}{4}$ . Of these four solution pairs, only $\frac{-1 \pm \sqrt{5}}{2}$ and $\frac{-1 \pm \sqrt{17}}{4}$ satisfy the condition of the problem.

The answer to this question is therefore $1+5+2+17+4=\boxed{29}$

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Before the readers start accusing me of pulling random Mathematical tricks out of nowhere, let's examine some of the observations which motivate this solution. First, notice that the coefficient of $z^{8}$ matches the constant, and the coefficient of $z^{7}$ nearly matches that of $z^{1}$ , bar the signage. This suggests that we should group terms with similar coefficients together and factorize the coefficient.

The next step takes more mathematical maturity to execute. Now that we have group and factorize terms with similar coefficients, it seems that we have run into a dead end. In such situations, a good rule of thumb to follow is that of creating symmetry to simplify a problem. But how can we create symmetry from terms like $(z^{8}+1)$ and $(z^{7}-z)$ ? Upon closer inspection, the sum of powers of terms in each individual parentheses add up to $8$ , that is, they are in the form $z^{a} \pm z^{8-a}$ . If we factorize $z^{4}$ , we would get terms in the form $z^{a-4} \pm z^{4-a}$ , or $z^{b} \pm z^{-b}$ . We have successfully introduce symmetry into our problem!

In general, $z^{a} \pm z^{2b-a}= z^{b}(z^{a-b} \pm z^{b-a})$ . Remember this cool Mathematical trick!

Hopefully, by this stage, you realize how to proceed. It might take some algebraic fortitude to write everything in terms of $(z-\frac{1}{z})$ , but the equation simply cries to be written in this way. Some of the tell-tale signs include the factorization of $a^{3}-b^{3}$ as $(a-b)(a^{2}+ab+b^{2})$ and the fact that $ab=1$ so conveniently, which simplifies a lot of terms in expansions.

The actual question is much more merciful in that it actually suggests the candidates to write everything in terms of $(z-\frac{1}{z})$ . However, I contend that this suggestion is actually the crux to solving this problem and thus have withheld it while posting the question. It is more fun this way. :)

Given

\displaystyle 2z^8 -3z^7 - 12z^6 + 12z^5 + 22z^4 - 12z^3 - 12z^2 + 3z +2 = 0. \ \ \ \ \tag{0}

It is clear that $z \neq 0$ since $z = 0$ cannot satisfy the above. Hence dividing the equation through by $z^4$ and grouping the terms with the reciprocals of the terms of the same degree yield

\displaystyle 2\left(z^4 + \frac{1}{z^4}\right) - 3\left(z^3 - \frac{1}{z^3}\right) - 12\left(z^2 + \frac{1}{z^2}\right) + 12\left(z - \frac{1}{z}\right) + 22 = 0. \ \ \ \ \tag{1}

Let $\displaystyle x = z - \frac{1}{z}$ . Taking the squares, cubes and the fourth order on both sides yields

$\displaystyle z^2 + \frac{1}{z^2} = x^2 + 2 \quad \text{and} \quad z^3 - \frac{1}{z^3} = x^3 + 3x \quad \text{and} \quad z^4 + \frac{1}{z^4} = x^4 + 4x^2 + 2$

respectively. Plugging all these into $(1)$ yields

\displaystyle 2x^4 - 3x^3 - 4x^2 + 3x + 2 = 0.\ \tag{2}

Clearly, $x \neq 0$ . Dividing $(2)$ through by $x^2$ and grouping in the same fashion as the above yield \displaystyle 2\left(x^2 + \frac{1}{x^2}\right) - 3\left(x - \frac{1}{x}\right) - 4 = 0.\ \tag{3}

Let $\displaystyle y = x - \frac{1}{x}$ . Then $\displaystyle x^2 + \frac{1}{x^2} = y^2 + 2$ so that $(3)$ becomes

$2y^2 - 3y = 0 \Longrightarrow y = 0 \ \text{or}\ y =\frac{3}{2}.$

$\textbf{Case 1:}$

For $y=0$ ,

$\displaystyle 0 = x - \frac{1}{x} \Longrightarrow x = \pm 1$ .

For $x = \pm 1$ ,

$\displaystyle \pm 1 = z - \frac{1}{z} \Longrightarrow z^2 \pm z - 1 = 0 \Longrightarrow z = \frac{1 \pm \sqrt{5}}{2} \ \text{or}\ \frac{-1 \pm \sqrt{5}}{2}.$

$\textbf{Case 2:}$

For $\displaystyle y=\frac{3}{2}$ ,

$\displaystyle 3 = 2x - \frac{2}{x} \Longrightarrow 2x^2 - 3x - 2=0 \Longrightarrow x = 2\ \text{or}\ -\frac{1}{2}$ .

For $x = 2$ ,

$\displaystyle 2 = z - \frac{1}{z} \Longrightarrow z^2 - 2 z - 1 = 0 \Longrightarrow z = 1 \pm \sqrt{2}$ .

For $x =-\frac{1}{2}$ ,

$\displaystyle -\frac{1}{2} = z - \frac{1}{z} \Longrightarrow 2z^2 + z - 2=0 \Longrightarrow z = \frac{-1 \pm \sqrt{17}}{4}$ .

Since $A$ must be positive, neither the roots $1 \pm \sqrt{2}$ nor $\frac{1 \pm \sqrt{5}}{2}$ are in the required form; only $\frac{-1 \pm \sqrt{17}}{4}$ and $\frac{-1 \pm \sqrt{5}}{2}$ satisfy the above condition. Then we have

$A = 1, B=5, C = 2, D = 17, E = 4$

and so

$A + B + C + D + E = 29$

as required.