1. There is a pair of four-digit numbers (2025, 3136) with the following characteristics: each number is a perfect square, and each digit of the second four-digit number is 1 greater than the corresponding digit of the first four-digit number. Please find all pairs of five-digit numbers that satisfy this characteristic. (If a pair of five-digit numbers found is $a$ and $b$, please write it in the form $(a, b)$)
【Analysis】Suppose we find the pair $\left(a^{2}, b^{2}\right)$, then we have:
$$
b^{2}-a^{2}=11111
$$

which means $(b-a)(b+a)=41 \times 271=1 \times 11111$. Considering $10000 \leq a^{2} \leq b^{2} \leq 99999 \Rightarrow 100 \leq a \leq b \leq 316$, the only solution is $\left\{\begin{array}{l}b-a=41 \\ b+a=271\end{array} \Rightarrow\left\{\begin{array}{l}a=115 \\ b=156\end{array}\right.\right.$
Thus, the only pair of numbers is $\left(115^{2}, 156^{2}\right)=(13225,24336)$