Theorem: Dudeney's Property of 2592

:$2592 = 2^5 \times 9^2$
It is the only number $n$ that has the property that:
:$n = \sqbrk {abcd} = a^b \times c^d$
where $\sqbrk {abcd}$ denotes the decimal representation of $n$.

Proof:
First we verify that $2592$ does indeed satisfy the given property.
{{begin-eqn}}
{{eqn | l = 2592
      | r = 2^5 \times 3^4
      | c = Prime Decomposition of $2592$
}}
{{eqn | r = 2^5 \times \paren {3^2}^2
      | c = 
}}
{{eqn | r = 2^5 \times 9^2
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
It remains to be shown that this is the only such number.
Because $\sqbrk {abcd} = a^b \times c^d$, neither $a^b$ nor $c^d$ can have more than $4$ digits.
Hence, for each digit, the highest power is:
{{begin-eqn}}
{{eqn | l = 1^9
      | r = 1
}}
{{eqn | l = 2^9
      | r = 512
}}
{{eqn | l = 3^8
      | r = 6561
}}
{{eqn | l = 4^6
      | r = 4096
}}
{{eqn | l = 5^5
      | r = 3125
}}
{{eqn | l = 6^5
      | r = 7776
}}
{{eqn | l = 7^4
      | r = 2401
}}
{{eqn | l = 8^4
      | r = 4096
}}
{{eqn | l = 9^4
      | r = 6561
}}
{{end-eqn}}
Neither $a$ or $c$ can be zero, or that would make $\sqbrk {abcd} = 0$.
Suppose $a = 1$ or $b = 0$.
Then:
:$\sqbrk {abcd} = c^d$
Apart from the above powers which have $4$ digits, we also have:
{{begin-eqn}}
{{eqn | l = 3^7
      | r = 2187
}}
{{eqn | l = 4^5
      | r = 1024
}}
{{eqn | l = 6^4
      | r = 1296
}}
{{end-eqn}}
Hence, by inspection, it is seen that none of these fit the pattern $\sqbrk {1bcd}$ or $\sqbrk {a0cd}$.
Similarly, suppose $c = 1$ or $d = 0$.
Then:
:$\sqbrk {abcd} = a^b$
Again, by inspection, it is seen that none of the above $4$-digit powers fit the pattern $\sqbrk {ab1d}$ or $\sqbrk {abc0}$.
Suppose either $a^b$ or $c^d$ has $4$ digits.
Then the other is less than $10$, giving:
{{begin-eqn}}
{{eqn | l = n^1
      | r = n
}}
{{eqn | l = 2^2
      | r = 4
}}
{{eqn | l = 2^3
      | r = 8
}}
{{eqn | l = 3^2
      | r = 9
}}
{{end-eqn}}
We try multiplying these by all the above $4$-digit powers such that their product is $4$ digits (there are not many).
First note that $1^1$ can be ruled out as none of these $4$-digit powers either begins or ends with $11$.
{{begin-eqn}}
{{eqn | l = 3^7 \times 2^1
      | r = 2187 \times 2
}}
{{eqn | r = 4374
}}
{{eqn | l = 3^7 \times 3^1
      | r = 2187 \times 3
}}
{{eqn | r = 6561
}}
{{eqn | l = 3^7 \times 4^1
      | r = 3^7 \times 2^2
}}
{{eqn | r = 2187 \times 4
}}
{{eqn | r = 8748
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 4^5 \times 2^1
      | r = 1024 \times 2
}}
{{eqn | r = 2048
}}
{{eqn | l = 4^5 \times 3^1
      | r = 1024 \times 3
}}
{{eqn | r = 3072
}}
{{eqn | l = 4^5 \times 4^1
      | r = 4^5 \times 2^2
}}
{{eqn | r = 1024 \times 4
}}
{{eqn | r = 4096
}}
{{eqn | l = 4^5 \times 5^1
      | r = 1024 \times 5
}}
{{eqn | r = 5120
}}
{{eqn | l = 4^5 \times 6^1
      | r = 1024 \times 6
}}
{{eqn | r = 6144
      | c = a near miss: $6^1 \times 4^5 = 6144$
}}
{{eqn | l = 4^5 \times 7^1
      | r = 1024 \times 7
}}
{{eqn | r = 7168
}}
{{eqn | l = 4^5 \times 8^1
      | r = 4^5 \times 2^3
}}
{{eqn | r = 1024 \times 8
}}
{{eqn | r = 8192
}}
{{eqn | l = 4^5 \times 9^1
      | r = 4^5 \times 3^2
}}
{{eqn | r = 1024 \times 9
}}
{{eqn | r = 9216
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 4^6 \times 2^1
      | r = 4096 \times 2
}}
{{eqn | r = 8192
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 5^5 \times 2^1
      | r = 3125 \times 2
}}
{{eqn | r = 6250
}}
{{eqn | l = 5^5 \times 3^1
      | r = 3125 \times 3
}}
{{eqn | r = 9375
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 6^4 \times 2^1
      | r = 1296 \times 2
}}
{{eqn | r = 2592
}}
{{eqn | l = 6^4 \times 3^1
      | r = 1296 \times 3
}}
{{eqn | r = 3888
}}
{{eqn | l = 6^4 \times 4^1
      | r = 6^4 \times 2^2
}}
{{eqn | r = 1296 \times 4
}}
{{eqn | r = 5184
}}
{{eqn | l = 6^4 \times 5^1
      | r = 1296 \times 5
}}
{{eqn | r = 6480
}}
{{eqn | l = 6^4 \times 6^1
      | r = 1296 \times 6
}}
{{eqn | r = 7776
}}
{{eqn | l = 6^4 \times 7^1
      | r = 1296 \times 7
}}
{{eqn | r = 9072
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 7^4 \times 2^1
      | r = 2401 \times 2
}}
{{eqn | r = 4802
}}
{{eqn | l = 7^4 \times 3^1
      | r = 2401 \times 3
}}
{{eqn | r = 7203
}}
{{eqn | l = 7^4 \times 4^1
      | r = 7^4 \times 2^2
}}
{{eqn | r = 2401 \times 4
}}
{{eqn | r = 9604
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 8^4 \times 2^1
      | r = 4096 \times 2
}}
{{eqn | r = 8192
}}
{{end-eqn}}
So none of the above $4$-digit powers multiplied by any of these single-digit powers satisfies the condition.
We have that $\sqbrk {abcd}$ cannot end in $0$, as that can also be ruled out by inspection of the $4$-digit powers.
Similarly, because $a \ne 1$, $\sqbrk {abcd} > 2000$.
Also, it cannot be the case that both $a^b$ and $c^d$ are smaller than $44$, as $44^2 < 2000$.
{{finish}}
{{Namedfor|Henry Ernest Dudeney|cat = Dudeney}}