A number with an interesting property. $abcd=a^b c^d$
We want to find integers $a,b,c,d$ such that $$1000a+100b+10c+d=a^bc^d\tag0$$ $$1\le a\le 9,\quad 0\le b\le 9,\quad 0\le c\le 9,\quad 0\le d\le 9$$
We use the followings (the proofs are written at the end of the answer) :
$(1)$ $c\not=0,1$
$(2)$ $d\not=0$
$(3)$ $b\not=0,1$
$(4)$ $a\not=1$
$(5)$ $d\not=1$
$(6)$ $c^d\le 2499$
$(7)$ If $d$ is odd, then $a,c$ have to be odd. If $d$ is even, then either $a$ or $c$ has to be even.
$(8)$ $(c,d)\equiv (0,0),(1,1),(1,2),(1,3),(2,0),(3,1),(3,2),(3,3)\pmod 4$
$(9)$ If $c=5$, then $d=5$.
$(10)$ If $a^b\equiv 2,3\pmod 5$ with $a$ even, then $a=2,8$.
$(11)$ If $a^b\equiv 2,3\pmod 5$ with $a$ odd, then $a=3,7$.
$(12)$ If $a^b\equiv 4\pmod 5$ with $a$ even, then $a=2,4,8$.
$(13)$ If $a^b\equiv 4\pmod 5$ with $a$ odd, then $a=3,7,9$.
From $(1)(2)(5)(6)(8)(9)$, we see that $(c,d)$ has to be either $$(4,4),(9,2),(9,3),(2,4),(2,8),(6,4),(3,2),(3,6),(7,2),(3,3),(3,7),(7,3),(3,5)$$ Here, let us separate it into cases and use $(3)(4)(7)(10)(11)(12)(13)$.
Case 1 : If $(c,d)=(4,4)$, then we have $250a+25b+11=64a^b$. So, $b$ has to be odd. Also, we have $4\equiv a^b4^4\pmod{10}\implies a^b\equiv 4,9\pmod{10}$ from which $(a,b)=(4,3),(4,5),(9,3)$ follow. However, these do not satisfy $a^b\le\frac{9999}{4^4}=39+\frac{15}{256}$.
Case 2 : If $(c,d)=(9,2)$, then we have $a+b+11\equiv a^b9^2\equiv 0\pmod 9\implies b\equiv 7-a\pmod 9$, and $a^b\equiv 2\pmod{10}$. Also, from $(4)(7)$, $a=2,8$. These imply $(a,b)=(2,5)$, and then the equation $(0)$ holds.
Case 3 : If $(c,d)=(9,3)$, then from $(7)(11)$, we have $a=3,7$. We also have $b\equiv 6-a\pmod 9$ from which $(a,b)=(3,3)$ follows. However, this does not satisfy $a^b\le\frac{9999}{9^3}=13+\frac{522}{729}$.
Case 4 : If $(c,d)=(2,4)$, then we have $250a+25b+6=4a^b$. So, $b$ has to be even. We also have $a^b\le\frac{9999}{2^4}=624+\frac{15}{16}$ and $a^b\equiv 4,9\pmod{10}$ from which $(a,b)=(2,2),(2,6),(3,2),(7,2),(8,2)$ follow. Also, we have to have $a+b\equiv a^b\pmod 3$ which does not hold for $(a,b)=(2,6),(3,2),(7,2)$. Finally, for $(a,b)=(2,2),(8,2)$, the equation $(0)$ does not hold.
Case 5 : If $(c,d)=(2,8)$, then we have $a^b\le\frac{9999}{2^8}=39+\frac{15}{256}$ and $a^b\equiv 3,8\pmod{10}$. These imply $(a,b)=(2,3)$ for which $a+b+1\equiv a^b\pmod 3$ does not hold.
Case 6 : If $(c,d)=(6,4)$, then we have $b\equiv 8-a\pmod 9$ and $a^b\equiv 4,9\pmod{10}$ from which $(a,b)=(2,6)$ follows. However, this does not satisfy $a^b\le\frac{9999}{6^4}=7+\frac{927}{1296}$.
Case 7 : If $(c,d)=(3,2)$, then, from $(7)(10)$, $a=2,8$. Also, we have $b\equiv 4-a\pmod 9$ from which $(a,b)=(2,2)$ follows. However, this does not satisfy $a^b\equiv 7\pmod{10}$.
Case 8 : If $(c,d)=(3,6)$, then, from $(7)(12)$, $a=2,4,8$. Also, we have $b\equiv 9-a\pmod 9$ and $a^b\equiv 4\pmod{10}$ from which $(a,b)=(4,5)$ follows. However, this does not satisfy $a^b\le\frac{9999}{3^6}=13+\frac{522}{729}$.
Case 9 : If $(c,d)=(7,2)$, then, from $(7)(10)$, $a=2,8$. Also, we have $a^b\equiv 8\pmod{10}$ from which $(a,b)=(2,3),(2,7)$ follow. However, for each case, the equation $(0)$ does not hold.
Case 10 : If $(c,d)=(3,3)$, then, from $(7)(13)$, $a=3,7,9$. Also, we have $b\equiv 12-a\pmod 9$ from which $(a,b)=(9,3)$ follows. However, this does not satisfy $a^b\le\frac{9999}{3^3}=370+\frac{1}{3}$.
Case 11 : If $(c,d)=(3,7)$, then we have $b\equiv 8-a\pmod 9$ and $a^b\equiv 1\pmod{10}$. There are no such $(a,b)$.
Case 12 : If $(c,d)=(7,3)$, then we have $a^b\equiv 1\pmod{10}$ from which $(a,b)=(3,4),(3,8),(7,4),(9,2),(9,4)$ follow. However, these do not satisfy $a^b\le\frac{9999}{7^3}=29+\frac{52}{343}$.
Case 13 : If $(c,d)=(3,5)$, then, from $(7)(11)$, $a=3,7$. Also, we have $b\equiv 10-a\pmod 9$ from which $(a,b)=(3,7),(7,3)$ follow. However, these do not satisfy $a^b\le\frac{9999}{3^5}=41+\frac{36}{243}$.
Therefore, $\color{red}{(a,b,c,d)=(2,5,9,2)}$ is the only solution.
Finally, let us prove $(1),(2),\cdots,(13)$.
$(1)$ $c\not=0,1$
Proof : If $c=0$, then $1000a+100b+1=0$ is not a four-digit number.
If $c=1$, then $a^{b-1}=1000+\frac{100b+10+d}{a}$ which implies $1000+\frac{10}{9}\le a^{b-1}\le 1919=1000+\frac{919}{1}$. So, $(a,b)=(4,6),(6,5)$. For each case, there is no such integer $d$. $\quad\blacksquare$
$(2)$ $d\not=0$
Proof : If $d=0$, then $1000a+100b+10c=a^b$ implies $10\mid a$ which is impossible. $\quad\blacksquare$
$(3)$ $b\not=0,1$
Proof : If $b=0$, then $1000a+10c+d=c^d$ implies $d\equiv c^d\pmod{10}$ which gives $(c,d)=(1,1),(7,3),(5,5),(4,6),(6,6),(3,7),(9,9)$. For each case, there is no such $a$.
If $b=1$, then $1000+\frac{100+10c+d}{a}=c^d$ which implies $1013+\frac{4}{9}=1000+\frac{121}{9}\le c^d\le 1000+\frac{199}{1}=1199$. So, we have $(c,d)=(4,5)$, but there is no such $a$. $\quad\blacksquare$
$(4)$ $a\not=1$
Proof : If $a=1$, then we have $1000+100b+10c+d=c^d$ which gives $1212\le c^d\le 1999$. So, we have $(c,d)=(6,4)$, but there is no such $b$. $\quad\blacksquare$
$(5)$ $d\not=1$
Proof : If $d=1$, then $a^{b-1}=\frac{1000a+100b+10c+1}{ca}$ which gives $27+\frac{34}{81}=\frac{2221}{9^2}\le a^{b-1}\le\frac{9991}{2^2}=2497+\frac 34$. So, $$\begin{align}(a,b)=&(2,6),(2,7),(2,8),(2,9),(3,5),(3,6),(3,7),(3,8),(4,4),(4,5),(4,6),\\&(5,4),(5,5),(6,3),(6,4),(6,5),(7,3),(7,4),(8,3),(8,4),(9,3),(9,4).\end{align}$$ For each case, there is no such integer $c$. $\quad\blacksquare$
$(6)$ $c^d\le 2499$
Proof : $c^d=\frac{1000a+100b+10c+d}{a^b}\le\frac{9999}{2^2}=2499+\frac 34$. $\quad\blacksquare$
$(7)$ If $d$ is odd, then $a,c$ have to be odd. If $d$ is even, then either $a$ or $c$ has to be even.
Proof : Since $b,d\not=0$, we have $a^bc^d\equiv ac\pmod 2$. So, $d\equiv 1000a+100b+10c+d\equiv a^bc^d\equiv ac\pmod 2$. $\quad\blacksquare$
$(8)$ $(c,d)\equiv (0,0),(1,1),(1,2),(1,3),(2,0),(3,1),(3,2),(3,3)\pmod 4$
Proof : We have $2c+d\equiv a^bc^d\pmod 4$. For $(c,d)\equiv (0,1),(0,2),(0,3)$, we have $d\equiv 0$ which is impossible. For $(c,d)\equiv (1,0),(3,0)$, we have $2\equiv a^b$ which is impossible since we already have $b\not=1$. For $(c,d)\equiv (2,2),(2,3)$, we have $2\ \text{or}\ 3\equiv 0$ which is impossible. Also, $(c,d)\equiv (2,1)$ is impossible from $(7)$. $\quad\blacksquare$
$(9)$ If $c=5$, then $d=5$.
Proof : We have $d\equiv a^bc^d\pmod 5$. If $c=5$, then $d\equiv 0\pmod 5$. $\quad\blacksquare$
$(10)$ If $a^b\equiv 2,3\pmod 5$ with $a$ even for some $b$, then $a=2,8$.
Proof : If $a=4$, then $a^b\equiv (-1)^b\equiv 1,4\pmod 5$ for any $b\ge 2$. If $a=6$, then $a^b\equiv 1\pmod 5$ for any $b\ge 2$. $\quad\blacksquare$
$(11)$ If $a^b\equiv 2,3\pmod 5$ with $a$ odd for some $b$, then $a=3,7$.
Proof : If $a=5$, then $a^b\equiv 0\pmod 5$ for any $b\ge 2$. If $a=9$, then $a^b\equiv (-1)^b\equiv 1,4\pmod 5$ for any $b\ge 2$. $\quad\blacksquare$
$(12)$ If $a^b\equiv 4\pmod 5$ with $a$ even for some $b$, then $a=2,4,8$.
Proof : If $a=6$, then $a^b\equiv 1\pmod 5$ for any $b\ge 2$. $\quad\blacksquare$
$(13)$ If $a^b\equiv 4\pmod 5$ with $a$ odd for some $b$, then $a=3,7,9$.
Proof : If $a=5$, then $a^b\equiv 0\pmod 5$ for any $b\ge 2$. $\quad\blacksquare$
I will type up the answer provided in the published solution for posterity.
From the American Mathematical Monthly problem E$69$ (Vol. $41$, May $1934$, p. $322$)
(accessed via JSTOR)
Instead of a product of powers, $a^bc^d$, a printer accidentally prints the four-digit number, $abcd$. The value is however the same. Find the number and show that it is unique.
- Since $abcd = a^bc^d$, neither $a^b$ nor $c^d$ can have more than four digits. Hence the highest possible powers for any single digits in the problem are $9^4$, $8^4$, $7^4$, $6^5$, $5^5$, $4^6$, $3^8$, $2^9$, and $1^9$ $2.$
- Neither $a$ nor $c$ is zero, as that would make $abcd$ zero
- If $a=1$, or if $b=0$, then $abcd = c^d$. Similarly, if $c=1$ or $d=0$, then $abcd = a^b$. Examination of the expanded powers which have four digits shows that none meets these conditions. Therefore $a \not =1$, $b \not = 0$, $ c \not = 1$, and $d \not =0$.
- If one of $a^b$ and $c^d$ has four digits, the other is less than ten and so must be $n^1$, $2^2$, $2^3$, or $3^2$. But no product of an eligible four digits expanded power by any one of these numbers meets condition (1), so these possibilities are accordingly ruled out.
- Since $abcd$ cannot end in zero, and since neither $a$ nor $c$ is one, $2000 \lt abcd \lt 10,000$.
- A test of the possible products formable from the remaining eligible expanded factors by multiplying the units digits and noting that the resultant units digits is not the same as either exponent, eliminates the majority of possibilities. Complete multiplication disposes of all others except $2^5 \cdot 9^2 = 32 \cdot 81 = 2592$, which is therefore the unique solution.