Given
$E \subseteq \mathbb {F}_q^d \times \mathbb {F}_q^d$
, with the finite field
$\mathbb {F}_q$
of order q and the integer
$d\,\ge \, 2$
, we define the two-parameter distance set
$\Delta _{d, d}(E)=\{(\|x-y\|, \|z-t\|) : (x, z), (y, t) \in E \}$
. Birklbauer and Iosevich [‘A two-parameter finite field Erdős–Falconer distance problem’, Bull. Hellenic Math. Soc.61 (2017), 21–30] proved that if
$|E| \gg q^{{(3d+1)}/{2}}$
, then
$ |\Delta _{d, d}(E)| = q^2$
. For
$d=2$
, they showed that if
$|E| \gg q^{{10}/{3}}$
, then
$ |\Delta _{2, 2}(E)| \gg q^2$
. In this paper, we give extensions and improvements of these results. Given the diagonal polynomial
$P(x)=\sum _{i=1}^da_ix_i^s\in \mathbb F_q[x_1,\ldots , x_d]$
, the distance induced by P over
$\mathbb {F}_q^d$
is
$\|x-y\|_s:=P(x-y)$
, with the corresponding distance set
$\Delta ^s_{d, d}(E)=\{(\|x-y\|_s, \|z-t\|_s) : (x, z), (y, t) \in E \}$
. We show that if
$|E| \gg q^{{(3d+1)}/{2}}$
, then
$ |\Delta _{d, d}^s(E)| \gg q^2$
. For
$d=2$
and the Euclidean distance, we improve the former result over prime fields by showing that
$ |\Delta _{2,2}(E)| \gg p^2$
for
$|E| \gg p^{{13}/{4}}$
.