Let F be a field of characteristic 2. Find the roots of $x^2+bx+c$ if we know $d$ such that $d(d+1)=b^{-2}c$.

Question

Let $F$ be a field of characteristic 2. Find the roots of $x^2+bx+c$ if we know $d$ such that $d(d+1)=b^{-2}c$, where ,b,c,d $\in F$.

I tried using the formula $\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-b\pm \sqrt{b^2-4c}}{2}$, and I am stumped by the fact that $2\equiv0 \mod 2$, so there will be division by $0$.

I also considered another method: To let $x+i$ and $x+j$ be the factor of $x^2+bx+c$. Then, $(x+i)(x+j)=x^2+(i+j)x+ij$, which implies $i+j=b$ and $ij=c$. But completely stumped as well?

How should I attempt this question? I have basic elementary group theory algebra knowledge.

Answer 1

If you multiply by $b^{-2}$ you get that $x^2+bx+c=0$ is equivalent to $(b^{-1}x)^2+b^{-1}x+d(d+1)=0$, which is equivalent to $b^{-1}x(b^{-1}x+1)=d(d+1)$. It therefore reduces to solving $y(y+1)=d(d+1)$ with $y=b^{-1}x$. Those solutions are $y=d$ and $y=d+1$, hence the original solutions are $x=bd$ and $x=b(d+1)$.

Answer 2

You have $ij=c=b^2d(d+1)$ so you need to find two elements $i$ and $j$ such that $$ i+j=b\qquad ij=(bd)(bd+b) $$ So you can take $i=bd$ and $j=bd+b$ (the order of the roots is irrelevant).

Since a polynomial of degree $2$ has at most two roots, these are the only roots.

Note that if $c\ne b^2d(d+1)$, for every $d\in F$, and $b\ne0$, then the polynomial has no roots in $F$. Indeed, we can reverse the argument. Say $bd$ and $be$ are the roots of the polynomial. Then $$ x^2+bx+c=(x+bd)(x+be)=x^2+b(d+e)x+b^2de $$ Therefore $d+e=1$ and so $e=d+1$, forcing $c=b^2d(d+1)$.