$10 \cdot 23$ has a four-digit number with thousands, hundreds, tens, and units digits being $a, b, c, d$ respectively, the sum of the digits is $26$, the tens digit of the product of $b, d$ equals $(a+c)$, and $(bd - c^2)$ is a power of 2. Find this four-digit number (explain the reasoning).
(Heilongjiang Province, China Mathematics Competition, 1979)
［Solution］From the problem, we have
$$
\left\{\begin{array}{l}
a+b+c+d=26, \\
b d=10(a+c)+e, \quad (e \text { is the unit digit of } b d) \\
b d-c^{2}=2^{m} \quad(m \in \mathbb{Z}) .
\end{array}\right.
$$

Since $\quad 0 \leqslant b \leqslant 9,0 \leqslant d \leqslant 9,0 \leqslant b d \leqslant 81$, we have
$$
\begin{array}{c}
1 \leqslant a+c \leqslant 8, \\
26=(a+c)+(b+d) \leqslant 8+(b+d),
\end{array}
$$

Thus,
$$
b+d \geqslant 26-8=18 .
$$

Also,
therefore
$$
\begin{array}{c}
b+d \leqslant 18, \\
b+d=18, \quad b=d=9 .
\end{array}
$$

And
$$
a+c=8 \text {. }
$$

From
$$
b d-c^{2}=2^{m} \text {, }
$$

we get
$$
(9+c)(9-c)=2^{m} \text {. }
$$

This equation holds only when $c=7$. Thus, $a=1$. Therefore, the required four-digit number is 1979.