Ratio of Number to Reversal which is Multiple

Theorem
Take a (strictly) positive integer $n$, written in conventional decimal notation.

Let $m$ be the reversal of $n$.

Let $m = k n$ where $k$ is an integer.

Then $k$ is either $4$ or $9$.

Existence
$8712 = 4 \times 2178$

$9801 = 9 \times 1089$

Uniqueness
Write $n = \sqbrk {a \dots b}$.

Then $m = \sqbrk {b \dots a} = k n$.

We must have $b = k a$ and $a \equiv k b \pmod {10}$.

Hence $a \equiv k^2 a \pmod {10}$ and $k a < 10$.

We only consider $2 \le k < 10$ since the case $k = 1$ is trivially satisfied by palindromes.

This leads to:
 * $1 \le a < 5$
 * $0 \equiv \paren {k^2 - 1} a \equiv a \paren {k - 1} \paren {k + 1} \pmod {10}$.

Hence one of $a$, $k - 1$, $k + 1$ is divisible by $5$.

By the restriction on $a$ and $k$, we have:
 * $5 \nmid a$
 * $k - 1 = 5$ or $k + 1 = 5$

Thus $k$ must be one of $4, 6, 9$.

Suppose $k = 6$.

From $k a < 10$, $a = 1$.

But $1 \paren {6 - 1} \paren {6 + 1} = 35$ is not divisible by $10$.

Hence $k$ must be $4$ or $9$.