Proving a mathematical curiosity.
Today, a thread full of cool math facts appeared on Reddit. As usual, someone mentioned the fact that 111111111 * 111111111 = 12345678987654321. In another reply, someone pointed out that this also works in other bases. For some reason, I decided that I needed to prove that it works in all bases.
To begin, I needed a general formula for values of the 111… terms. This was fairly straightforward: for a base $b$, we want $b-1$ base-$b$ digits, all ones. To standardize the base, we multiply each digit by an increasing power of $b$ and sum. Since each digit is one, we get a nice geometric series which can easily be solved.
$$ \sum\limits_{i=1}^{b-1} b^{i-1} = \frac{b-b^b}{b-b^2}$$
When we multiply this number by itself, we are squaring it, so we end up with $\left((b-b^b)/(b-b^2)\right)^2$.
The hard part was writing a general form for the $1234 \dots (b-1) \dots 4321$ number. To deal with this, I broke it down into two parts, as illustrated below.
| Digit value | 1 | 2 | $(b-2)$ | $(b-1)$ | $(b-2)$ | 2 | 1 | ||
| $b^{2b-3}$ | $b^{2b-4}$ | $\dots$ | $b^{2b-(b+1)}$ | $b^{2b-(b+2)}$ | $\dots$ | ||||
| $b^{b-2}$ | $b^{b-3}$ | $b^1$ | $b^0$ |
I calculated the values of the most-significant digits starting at the left, and the values of the least-significant digits starting at a right. To make the math come out nicely, I actually included the center digit in both formulas. That’s okay, since we can subtract it off once to make up for the duplicate. Now we have a summation formula for the value of the square.
$$ \left( \sum\limits_{i=1}^{b-1} ib^{2b-(i+3)} \right) + \left( \sum\limits_{i=1}^{b-1} ib^{i-1} \right) - b^{b-2} $$
With a little thinking (or the help of a computer algebra system), we can get a neat closed form.
$$ \left( \sum\limits_{i=1}^{b-1} ib^{2b-(i+3)} + ib^{i-1} \right) - b^{b-2} = \left(\frac{b-b^b}{b^2-b}\right)^2 $$
We can see that this is quite similar to the expression we got for the square above; the only difference is that the $b-b^2$ denominator has changed to $b^2-b$. Fortunately, this negation goes away when squaring, so we can trivially prove that the two expressions are equal.
And there we have it: proof that this curiosity is true in any base of at least two.