Consider the two symmetric matrices $$ \begin{pmatrix} 1 & a \\ a & 1 \end{pmatrix}$$ and $$ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ over the field $F_{11}$. For which values of $a$ are the associated quadratic spaces isomorphic?
I've tried the basic and brutal way to check the existence of the map but failed. It seems that this question has some trick I don't know. Is there any property that may be helpful?
Assuming that you know what the determinant of a non degenerate quadratic form is, it is easy to see that if $det(q)=dK^{\times 2}\in K^\times/K^{\times 2}$, and if $q$ represents a nonzero value $u\in K^\times$, then $q\simeq \langle u, ud\rangle$ (pick $v$ such tht $q(v)=u$, then complete $v$ to a $q$-orthogonal basis, and use the fact that the determinant is an invariant of the isomorphis class)
In particular, two non degenerate quadratic forms of rank $2$ are isomorphic if and only if they represents a common nonzero value and have same determinant.
Now both your quadratic forms represent $1$. The determinant of the first one is $1-a^2$ mod squares, and the determinant of the second one is $1$.
Hence your two quadratic forms are isomrophic if and only if $1-a^2$ is a square in $\mathbb{F}_{11}$. I let you find the values of $a$ which work (note that $a=1$ does not work since you want a non degenerate form, as the second one is)