$F(\alpha, \beta):F$ is a simple extension Let $K:F$ be a extension, $F$ characteristic other than $2$, $\alpha, \beta \in K-F$ with $\alpha \neq \beta$ and $\alpha \neq -\beta$ such that $\alpha^2, \beta^2 \in F$. I have to prove extension $F(\alpha, \beta):F$ is simple.
So as $F$ characteristic other than $n$, then $n(1)=0$ in $F$, $n \neq 2$.
We want to show that there exist $x$ in $F(\alpha, \beta)$ such that $F(\alpha, \beta)=F(x)$.
What can I do next?
Can somebody help me with a hint or solution ?
 A: As with the hint of @Chickenmancer, consider $F(a+b)$. Then clearly $F(a+b) \subset F(a, b)$. We'll show we have the reverse inclusion and hence equality. 
Since $(a+b) \in F(a+b)$ we must have that $(a+b)^2 = a^2 + 2ab + b^2 \in F(a+b)$.  
Now, $a^2, b^2 \in F \subset F(a+b)$ gives us that $2ab \in F(a+b)$.
Now the characteristic isn't 2, so we can divide by the $2$ since we're in a field so $ab \in F(a+b)$.
Then, since $a \ne -b$ we have $a+b \ne 0$ and so we can divide by $(a+b)$ in $F(a+b)$.  
Now we see that, $(a-b) = \frac{(a-b)(a+b)}{a+b}= \frac{a^2 - b^2}{a+b} \in F(a+b)$.  
Lastly $(a-b) + (a+b) = 2a \in F(a+b)$. Again by characteristic not equal to $2$ we get that $a \in F(a+b)$. Similarly $b \in F(a+b)$. So $F(a, b) = F(a+b)$ and we're done. 
A: It should be clear that $F(\alpha + \beta) \subseteq F(\alpha ,\beta)$.
Note that in $F(\alpha, \beta)$, we can compute the inverse of $\alpha + \beta$:
$$ \frac{1}{\alpha + \beta} = \frac{1}{\alpha + \beta}\left( \frac{\alpha - \beta}{\alpha - \beta}\right) = \frac{ \alpha - \beta}{\alpha^2 - \beta^2}.$$
Solving for $\alpha - \beta$ gives:
$$ \alpha - \beta = \frac{\alpha^2 - \beta^2}{\alpha + \beta}.$$
But $\alpha^2 - \beta^2 \in F$ and $\frac{1}{\alpha + \beta} \in F(\alpha + \beta)$. Thus $\alpha - \beta \in F(\alpha + \beta)$.
Can you use this to show that $\alpha , \beta \in F(\alpha + \beta)$?
