Which coefficient is greater?

Question

If $b, c$ are integers, and $\sqrt{2} + \sqrt{3}$ is a root of the equation, $x^4 + bx^2 + c = 0$, which is greater, $b$ or $c$; where $b, c$ are both integers.

Since $\sqrt{2} + \sqrt{3}$ is a root, sub it in gives:

$ (5 + 2\sqrt{6})b + c = -49 - 20\sqrt{6}$

This gives:

$$c = -49 - 20\sqrt{6} - (5 + 2\sqrt{6})b$$

$$b = \frac{-49 - 20\sqrt{6} - c}{5 + 2\sqrt{6}}$$

If $c > 0$ then $b < 0$ and if $b > 0$ then $ c < 0$.

So how to go about this?

I know that $\sum \text{roots} = 0$ hence, WLOG, $r_2 + r_3 + r_4 = -\sqrt{2} - \sqrt{3}$, but that doesn't help.

Answer 1

$$\begin{align}x=\sqrt 2+\sqrt 3&\Rightarrow x-\sqrt 2=\sqrt 3\\&\Rightarrow x^2-2\sqrt 2x+2=3\\&\Rightarrow 2\sqrt 2x=x^2-1\\&\Rightarrow 8x^2=x^4-2x^2+1\\&\Rightarrow x^4-10x^2+1=0\end{align}$$

So, $x^4-10x^2+1=0$ is an equation which has a root $x=\sqrt 2+\sqrt 3$.

Suppose that $x^4+px^2+q=0$ has a root $x=\sqrt 2+\sqrt 3$ where $p,q$ are integers such that $(p,q)\not=(-10,1)$. Then, we have $$(x^2=)\ 5+2\sqrt 6=\frac{q-1}{-10-p}.$$ The LHS is irrational, and the RHS is rational, which is a contradiction.

So, we have $b=-10,c=1$.

Answer 2

One way is to notice that the four roots must be $\sqrt 2+\sqrt 3$, $\sqrt2-\sqrt 3$, $-\sqrt 2+\sqrt 3$, $-\sqrt 2-\sqrt 3$.

Another (faster and simpler, I suppose) way is to notice that we are looking for a quadratic equation $y^2+by+c=0$ where $(\sqrt 2+\sqrt 3)^2=5+2\sqrt 6$ is a root and hence $5-2\sqrt 6$ must be the other.