Suppose that $F$ is a field. Show that there exists a $F$-division algebra $D$ with two elements $a\neq b\in D$ such that $a^2-2ab+b^2=0$.


In the field extensions we know that $a^2-2ab+b^2=0$ if and only if $a=b$, because of $a^2-2ab+b^2=(a-b)^2$. But I know this is not true if the extension is division ring, but I can't prove it.


In a part of my research I want two elements like that in order to extend $F$ to $F(a)$ such that there exists $b\notin F(a)$ such that we know its minimal polynomial $p(x)=x^2-2ax+a^2$. 


It is a little part of my research and I don't have any idea to prove or simplify this question. About 3 years ago one of my professors told me this is true but now, he isn't in touch.



Division ring on a field
I assume $\text{char}\,\mathbf F=0$.


Put $d:=b-a$. Because of $a^2-2ab+b^2=d^2-ad+da$ your equation is equivalent to
$$
(\*)\qquad d^{-1}a-ad^{-1}=1.
$$
This precludes $\dim\_{\mathbf F}D<\infty$ (take the reduced trace on both sides).


On the other side, $(\*)$ is is the relation defining of the Weyl algebra $A\_1(\mathbf F)=\mathbf F\langle x,\partial\_x\rangle$ and it is well known that $A\_1$ has a skew field of fractions $D$. Then
$$
a=x,b=x+\partial\_x^{-1}\in D
$$
is a solution to your problem. Another would be
$$
a=\partial\_x, b=\partial\_x-x^{-1}.
$$


