Prove that the polynomial $f(x)= x^4 + 2x^2 + 2x + 1998$ cannot be writen as the product of two quadratic polynomials with integer coefficients

Question

Assume that $f(x)=p(x)q(x)$ with $p(x),~q(x)$ quadratics over $\mathbb{Z}$. Comparing coefficients of $x^4$, the leading coefficients of $p$ and $q$ are both equal to $+1$ or $-1$. In the second case, one can replace the quadratics by their negatives. Then one can write

$$p(x)=x^2+ax+b, \quad q(x)=x^2+cx+d,$$

with $a,~b,~c,~d\in\mathbb{Z}$. Doing the calculations and comparing coefficients gives the nonlinear system

$$a+c=0,\quad ac+b+d=2,\quad ad+bc=2,\quad bd=1998.$$

From the first equation, $c=-a$. Replacing $c=-a$ into the second one gives $d=a^2-b+2$. Replacing $c=-a$ and $d=a^2-b+2$ into the third equation gives

$$a(a^2-2b+2)=2$$

I am puzzled how to use this equation and the last one of the system, $bd=1998$, to get a contradiction. Thanks in advance for any help.

Answer 1

Use only $a+c=o, ad+bc=2\Leftrightarrow a(d-b)=2\Rightarrow d=b\pm1, \pm2$ while bd=1998. Then try to solve the resulting quadratic in b.

Answer 2

Easiest way: use a calculator like Wolfram Alpha to compute the four roots, then check every possible product $(x-\alpha)(x-\beta)$, where $\alpha$ and $\beta$ are roots.