Check equivalence of quadratic forms over finite fields How to check whether the two quadratic forms \begin{equation} x_1^2 + x_2^2 \quad \text{(I)}\end{equation} and 
\begin{equation} 2x_1x_2 \quad \text{(II)} \end{equation} 
are equivalent on each of the spaces $\mathbb{F}_3^2\, \text{and}\,\mathbb{F}_5^2$? 
I know that these forms correspond to the two matrices \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} in case (I) and 
\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} in case (II). But what do I do with those and what difference do the different fields make? 
 A: You've wrtten the other way round. For the quadratic form $ x_1^2 + x_2^2 $, the corresponding matrix is the identity while that for $ 2x_1x_2 $ is $ A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $. The forms are equivalent iff the matrices are congruent. (over $ \mathbb{F}_3 $ and $ \mathbb{F}_5 $) 
So, suppose $ A = P^TIP = P^TP $ over $ \mathbb{F}_3 $ for some invertible $ P $. Then, taking determinants gives $ -1 = ( \det P)^2 $, which is a contradiction as $ -1 $ is not a quadratic residue modulo $ 3 $.
Over $ \mathbb{F}_5 $, you can check that they indeed are equivalent by the matrix $ P = \begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix} $ as we have, $ (3x_1+x_2)^2 + (x_1+3x_2)^2 = 2x_1x_2 $ in $ \mathbb{F}_5 $.
A: The characteristic of the field makes a big difference in the computations. I would start by comparing the discriminates of the two matrices.
Over both fields, the discriminants are $\pm 1$. The question is: do $\pm1$ differ by a square factor in both fields or not?
Over the field of five elements, $1\cdot 3^2\equiv -1\pmod 5$, but over the field of three elements, $a^2\equiv 1\not\equiv -1\pmod 3$ for all nonzero $a$.
After some work, you can find that the two matrices are cogredient this way: $$\begin{bmatrix}3&1\\1&3\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}3&1\\1&3\end{bmatrix}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$$ 
Here's an elementary brute force way to figure this out using just algebra. From $$
\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}a&c\\b&d\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}a&c\\b&d\end{bmatrix}=\begin{bmatrix}a^2+b^2&ac+bd\\ac+bd&c^2+d^2\end{bmatrix}=\begin{bmatrix}0&1\\1&0\end{bmatrix}
$$
we have already learned that $a^2=-b^2$, $c^2=-d^2$ and $ac+bd=1$. Earlier I noted that $1$ differs from $-1$ by a factor of $3^2$, so I suppose my change of basis matrix has determinant $3$, so $ad-bc=3$.
Now 
$$
a= a^2c+adb\\ 
=-b^2c+adb\\
= b(ad-bc)\\
= 3b
 $$
Similarly $$
c= ac^2+dcb\\ 
=-d^2a+dcb\\
= -d(ad-bc)\\
= -3d = 2d
 $$
We've consumed the two equations involving squares, and now we combine the other two.
We have that 
$$
1=ac+bd=6bd+bd=7bd=2bd
$$
Hence, $bd=3$ or in other words $d=3/b$. At this point we have $a,c,d$ all in terms of $b$ ($a=3b$, $d=3/b$, $c=2d=1/b$).
Choosing $b=1$, we arrive at $a=d=3$, $b=c=1$.
