Prove these quadractic forms are equivalent over $\mathbb{Z_5}$

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Consider the following quadractic forms, defined in the field $\mathbb{Z_5}$,

$$q(x, y, z, t) = 2y^2 + z^2 + 2t^2 + 4xy + 2xt + 4yt$$ $$q_0(x, y, z, t) = x^2 + y^2 + z^2 + dt^2$$

Prove they are equivalent for some $d\in\mathbb{Z_5}$.

Now, what I've done is, I've diagonalized the first quadractic form, which yields $diag(1,-2,2,\frac{1}{2})$. The second one is already in diagonal form, since its matrix is $diag(1,1,1,d)$.

To simplify things, the matrix of the first quadratic form can be simplied by the congruent matrix $diag(1,2,2,3)$.

Now, to prove they are equivalent, I'd have to show the matrices are congruent for some value $d\in\mathbb{Z_5}$, but I have no clue how to do this. Could you provide me with some suggestions?

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Hint: $$ \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}^T \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix}2 & 0 \\ 0 & 2 \end{pmatrix}$$