Quartic with no rational solutions

Let $f(x) = x^4-2x^2-8x-3$ . If $f(x)$ can be factorized into product of polynomials with rational coefficients, it could either be one linear factor with a cubic factor or two quadratic factors. We note that $f(\pm1) \ne 0$ , and $f(\pm3) \ne 0$ . By rational root theorem, this means that $f(x)$ has no linear factor and therefore it is of the form $(x^2 + Ax + C) (x^2+Bx+D)$

$\begin{aligned} \text{Assume } x^4-2x^2-8x-3 & = (x^2 + Ax \pm 1) (x^2+Bx \mp3) \\ & = x^4 + (A+B)x^3 + (AB \pm 1 \mp 3)x^2 + (\mp 3A + \pm B)x -3 \end{aligned}$

Equating coefficients on both sides, we have:

\(\begin{array} {} A+B = 0 & \Rightarrow B = -A \\ AB = -A^2 & = \begin{cases} -1+3-2 = 0 & \Rightarrow A = 0 & \color{red}{\text{rejected}} \\ 1-3-2 = -4 & \Rightarrow A = \pm 2 & \color{blue}{\text{accepted}} \end{cases} \end{array} \)

\(\begin{array} {} \Rightarrow A = \pm 2 \quad B = \mp 2 \quad C = -1 \quad D = 3 \end{array} \)

$\begin{aligned} \text{And note that } x^4-2x^2-8x-3 & = (x^2 + 2x -1) (x^2-2x+3) = 0 \end{aligned}$

And the roots are: $x = \begin{cases} \dfrac{-2 \pm \sqrt{4+4}}{2} & = \begin{cases} a = -1 + \sqrt{2} & \Rightarrow |a| = \sqrt{2} - 1 \\ b = -1 - \sqrt{2} & \Rightarrow |b| = 1 + \sqrt{2} \end{cases} \\ \dfrac{2 \pm \sqrt{4-12}}{2} & = \begin{cases} c = 1 + i\sqrt{2} & \Rightarrow |c| = \sqrt{3} \\ d = 1 - i\sqrt{2} & \Rightarrow |d| = \sqrt{3} \end{cases} \end{cases}$

Therefore, $\quad |a|+|b|+|c|+|d| = 2\sqrt{2} + 2\sqrt{3}\quad \Rightarrow m^2 + n^2 = \boxed{13}$

I started writing the polynomial as a product $(x^2 - bx+c_1)(x^2+bx+c_2),$ realizing that the linear terms should be opposites because the cubic term vanishes. It would be really nice if $c_1$ and $c_2$ are integers; so I factored the constant term as $(-1)\cdot 3$ or $(-3)\cdot 1$ . But we don't want $c_1+c_2 = -2$ , because then the contribution $-bx^2$ to the quadratic term would be zero, and there would be no linear term. Therefore I chose $c_1 + c_2 = 2$ , i.e. $c_1 = -1, c_2 = 3$ . It was now easy to decide on $b = 2$ , so that the equation factors as $(x^2 - 2x - 1)(x^2 + 2x + 3) = 0.$ The solutions are now easily found to be $1\pm\sqrt 2$ and $-1\pm\sqrt 2 i$ , with absolute values $\sqrt 2\pm 1$ and $\sqrt 3$ (twice). Therefore $|a|+|b|+|c|+|d| = 2\sqrt 2 + 2\sqrt 3,$ and the solution is $2^2 + 3^2 = \boxed{13}$ .

by the note ,we are to find that the cubic is $q(z)=z^3-\frac{2}{2}z^2+\frac{2^2+4*3}{16}z-\frac{8^2}{64}=z^3-z^2+z-1$ we see that $a_2=-8$ which is negative telling us that we are looking for positive square root sum of q's roots. $z^3-z^2+z-1=z^2(z-1)+1(z-1)=(z-1)(z^2+1)=(z-1)(z+i)(z-i)$ so the roots are 1,-i,i. we find that $x=\sqrt{1}+\sqrt{i}+\sqrt{-i}=1+\sqrt{2}$ we put this value and get zero.we know that in most polynomails, if $k+\sqrt{s}$ is a root, then $k-\sqrt{s}$ is another root. we put in x= $1-\sqrt{2}$ and also get zero. next we divide $\frac{x^4-2x^2-8x-3}{(x-(1+\sqrt{2}))(x-(1-\sqrt{2}))}=\frac{x^4-2x^2-8x-3}{x^2-2x-1}=x^2+2x+3$ we factor that part to get the othe two roots $-1-i\sqrt{2},-1+i\sqrt{2}$ . you can do the rest