Dudeney's Property of 2592

Theorem

 * $2592 = 2^5 \times 9^2$

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.

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:

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:

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:

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 $11$.

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 $44^2 < 2000$.