1. There are two four-digit numbers, their difference is 534, and the last four digits of their square numbers are the same. Then the larger four-digit number has ( ) possibilities.
(A) 1
(B) 2
(C) 3
(D) 4
- 1.C.

Let the larger four-digit number be $x$, and the smaller four-digit number be $y$. Then,
$$
x-y=534 \text {, }
$$

and $x^{2}-y^{2}$ is divisible by 10000.
Since $x^{2}-y^{2}=(x+y)(x-y)=267 \times 2(x+y)$, then $x+y$ is divisible by 5000.
Let $x+y=5000 k\left(k \in \mathbf{N}_{+}\right)$.
Solving equations (1) and (2) yields
$$
\left\{\begin{array}{l}
x=2500 k+267, \\
y=2500 k-267 .
\end{array}\right.
$$

Considering that $x$ and $y$ are both four-digit numbers, we have
$$
\left\{\begin{array}{l}
1000 \leqslant 2500 k+267 \leqslant 9999, \\
1000 \leqslant 2500 k-267 \leqslant 9999 .
\end{array}\right.
$$

Solving this gives $\frac{1}{2} \frac{267}{500} \leqslant k \leqslant 3 \frac{558}{625}$.
$k$ can take the values 1, 2, or 3.
Thus, $x$ can take three values: 2767, 5267, 7767.