There is a four-digit number $\overline{a b c d}, a+b+c+d=26, b \cdot d$'s tens digit equals $a+c$, and $b d-c^{2}$ is an integer power of 2. Try to find this four-digit number (explain the reasoning).
Parse Yan $0 \leqslant b \cdot d \leqslant 9 \cdot 9=81$, so $1 \leqslant a+c \leqslant 8$. Thus $26=(b+d)+$ $(a+c) \leqslant(b+d)+8$, which means $b+d \geqslant 18$. Therefore, $b=d=9, a+c=8$. Since $2^{m}=b d-c^{2}=$ $(9+c)(9-c)$, we have $c=7$, thus $a \doteq 1$. Hence, the required four-digit number is 1979.