elements in Field such that $1+\alpha a^2+\beta b^2=0$

Question

If $F$ is a finite filed, $\alpha, \beta$ are two non-zero elements of $F$. Then Show that there exists elements $a,b$ in $F$ such that $1+\alpha a^2+\beta b^2=0$.

I don't have any clue on how to approach this. Please help me with this problem.

Answer 1

The equation is equivalent to $$1+\alpha a^2=-\beta b^2.$$ If more than half the elements in the finite field $F$ are squares, then $1+\alpha a^2$ takes more than half the values in $F$ and $-\beta b^2$ takes more than half the values in $F$. So there has to be a collision: some value of $1+\alpha a^2$ must equal some value of $-\beta b^2$.

If $F$ has even order, every element of $F$ is a square.

If $F$ has odd order, it has $\frac12(1+|F|)$ squares.