A number that is its own square

Numbers that wrap set up arithmetic on a dial: add and multiply as usual, then keep only the remainder. On a dial of $10^n$, that means keeping the last $n$ digits. Arithmetic with the last few digits of numbers is arithmetic on such a dial, and every schoolchild who has noticed that the square of a number ending in 5 ends in 25 has been doing it.

Some numbers survive squaring with their last digits intact. $5^2 = 25$ ends in 5; $25^2 = 625$ ends in 25; $625^2 = 390625$ ends in 625. The same happens on the other side: $6^2 = 36$ ends in 6, $76^2 = 5776$ ends in 76, $376^2 = 141376$ ends in 376, $9376^2 = 87909376$ ends in 9376. This essay follows those two families leftward, digit by digit, until they become two infinite strings, and finds that they are a picture of the most useful theorem about dials.

Four numbers that are their own squares #

On a dial of $10^n$, the equation $x^2 = x$ asks for numbers whose square ends in the number itself. It has the obvious solutions $0$ and $1$, and for every $n$ it has exactly two more.

The reason is two dials at once. The Chinese remainder theorem says that a number on the dial of $10^n = 2^n \cdot 5^n$ is the same thing as a pair of numbers, one on the dial of $2^n$ and one on the dial of $5^n$, and that adding or multiplying on the big dial is adding or multiplying each member of the pair separately. So $x^2 = x$ on the big dial means $x^2 = x$ on both small dials at once.

On the dial of a prime power the equation has only the obvious solutions. It says $x(x - 1) = 0$, and $x$ and $x - 1$ cannot both be divisible by the prime, since they differ by one; so one of them carries every factor of $2^n$, or every factor of $5^n$, and $x$ is $0$ or $1$ there. With two dials and two choices on each, there are four solutions on the big dial: $(0, 0)$, which is $0$; $(1, 1)$, which is $1$; and the two mixed ones. For four digits these are $0625$, which is $0$ on the dial of 625 and $1$ on the dial of 16, and $9376$, which is the reverse.

The digits never change once set #

Going from $n$ digits to $n + 1$, the mixed solutions do not jump about: the new solution ends in the old one. The solution that is $0$ modulo $5^{n+1}$ and $1$ modulo $2^{n+1}$ is in particular $0$ modulo $5^n$ and $1$ modulo $2^n$, so its last $n$ digits are the old solution. Each solution extends the previous one by one digit on the left, and the digits, once written, are permanent.

The two families therefore define two infinite strings of digits running leftward:

$$\ldots 212890625 \qquad\text{and}\qquad \ldots 787109376.$$

Neither is a whole number — a whole number has finitely many digits — but both are ten-adic numbers, the base-ten counterpart of the seven-adic numbers a root lifted one digit at a time built for the square root of two. A ten-adic number is a consistent choice of last digits for every length, and arithmetic on them is arithmetic on all the dials at once: to multiply two of them to $n$ digits, multiply their last $n$ digits and keep the last $n$ digits of the answer.

In that arithmetic, each of the two strings is its own square: squaring changes none of its digits, at any length. Both were reached in the hero figure by repetition, and the way they are reached says something precise about the two primes.

One digit a step, none, or twice as many #

Squaring $5$ again and again gives $5^{2^k}$. Its distance from the limit is about whether $5^{2^k}$ is $1$ modulo a high power of two, since it is already $0$ modulo every power of five it needs. And $5^{2^k} - 1$ is divisible by exactly $2^{k+2}$: each squaring adds one factor of two, by the identity $5^{2m} - 1 = (5^m - 1)(5^m + 1)$, in which the second factor is twice an odd number. So each squaring settles exactly one more decimal digit.

Squaring $6$ does not work at all, which is the surprise in the figure. The powers $6, 36, 1296, 1679616, \ldots$ end in $6$ every time and never settle a second digit. The trouble is on the dial of five: $6^{2^k} - 1$ is divisible by $5$ exactly once, however many times $6$ is squared, because squaring adds factors of two and the obstacle is a factor of five. What works is raising to the fifth power. $6^{5^k} - 1$ is divisible by exactly $5^{k+1}$, by the same identity with five factors instead of two, and the hero figure’s lower family settles one digit at each fifth power.

The asymmetry is a small instance of a general rule about how powers approach $1$ on a prime’s dial — the exponent must keep gaining factors of that prime — and it shows the two dials pulling separately. Each limit is $1$ on one dial and $0$ on the other, and each iteration has to make progress on the dial where it is $1$.

Newton’s method is the efficient alternative. Applied to $f(x) = x^2 - x$ it replaces $x$ by $x - f(x)/f'(x) = x^2/(2x - 1)$, and on these dials, as on the dials of a prime in a root lifted one digit at a time, it doubles the number of settled digits at every step: $1, 2, 4, 8, 16, 32, \ldots$, reaching 256 digits in eight steps. The division by $2x - 1$ is possible because $2x - 1$ is $\pm 1$ on each small dial, so it has an inverse on the big one.

Why the powers must keep gaining factors #

The rule that decided which iterations converge can be stated once and used everywhere. On the dial of a prime power $p^n$, a number $a$ that is $1$ modulo $p$ — like $5$ modulo $2$, or $6$ modulo $5$ — has powers that drift towards $1$ only as their exponent gains factors of $p$. Precisely, the power of $p$ dividing $a^m - 1$ is the power dividing $a - 1$, plus the power of $p$ dividing $m$ — for $p = 2$ with a small correction when $m$ is even, which is why the count for $5$ starts at $k + 2$ rather than $k + 1$. This “lifting the exponent” rule is a cousin of the digit-by-digit lifting of a root lifted one digit at a time: there a root was improved by one digit for each correction, here a power is improved by one digit for each factor of $p$ in its exponent.

So the exponents $2^k$ gain factors of two and nothing else, which is why squaring helps $5$, whose obstacle is the dial of two, and does not help $6$, whose obstacle is the dial of five. The exponents $5^k$ gain factors of five, which is why fifth powers rescue $6$. An exponent that gains both — $10^k$ — would serve both strings at once: $5^{10^k}$ and $6^{10^k}$ converge as well, settling a digit a step, since the factor of two serves the first and the factor of five the second.

The iterations are fixed-point iterations of the kind the staircase that shows the whole orbit drew for real numbers, with a different notion of nearness. Squaring is the map $x \mapsto x^2$, and its fixed points on the ten-adic numbers are exactly the four solutions. Whether an orbit approaches a fixed point depends on the map’s slope there, measured in the adic sense: at $\ldots 890625$ the slope $2x$ is $2$ times a number that is $1$ on the dial of two, so it carries one factor of two, and the orbit approaches at a rate of one binary digit per step on that dial — which, because the dial of five is already satisfied, shows up as one decimal digit per step. At $\ldots 109376$ the slope is $2x$ again, with no factor of five, and on the dial of five the orbit makes no progress at all.

A powers table that never closes #

The same splitting explains a fact about last digits that people meet without explanation. The last digit of $n^5$ is always the last digit of $n$, for every $n$: $2^5 = 32$, $3^5 = 243$, $7^5 = 16807$. On the dial of two, every number is its own fifth power because there are only two numbers; on the dial of five, every number is its own fifth power by Fermat’s little theorem, which every element is a power of one of them placed inside the structure of the dial’s multiplication. Two dials, each satisfied, make the dial of ten satisfied.

The automorphic strings are the extreme case of this. The last digit of $n^5$ equals that of $n$ for every $n$, but the last two digits of $n^5$ equal those of $n$ only for some, and the last $k$ digits of $n^{5^k}$, or of $n^{2^k}$, settle only for $n$ that are $0$ or $1$ on each dial. The table of last digits of powers, extended to more and more digits, is a table that never closes up — except along the four rows that are the four ten-adic solutions of $x^2 = x$, where it closes at every length at once.

Two numbers that multiply to nothing #

The two strings add up to one and multiply to nought.

The sum is clear from the dials: $(1, 0) + (0, 1) = (1, 1)$, which is $1$. The product is $(1, 0) \times (0, 1) = (0, 0)$, which is $0$. In the digits, the product of $\ldots 109376$ and $\ldots 890625$ has every last digit nought, to any length one cares to compute, although neither factor ends in a single nought.

In the ordinary integers a product is nought only if a factor is. Numbers on the dial of a prime keep that property, and so do the $p$-adic numbers for a prime $p$; they form a number system in which division by anything non-zero is possible, which is what made the seven-adic square root of two behave like a real number. The ten-adic numbers do not. They have zero divisors, non-zero numbers whose product is nought, and so they are not a field in the sense of the field with four elements, where every non-zero element can be divided by, and the familiar argument “if $ab = 0$ then $a = 0$ or $b = 0$” fails in them. It is the failure that occurs on every dial whose size is not a power of a single prime — on the dial of six, $2 \times 3 = 0$ — and that multiplying every number on the dial at once ran into when multiplication by a number sharing a factor with the dial failed to rearrange it; here it is carried out to infinitely many digits.

The two strings make the failure exact. Every ten-adic number $x$ splits as $x = x\,e + x\,(1 - e)$, where $e = \ldots 109376$, and the two parts are its base-two and base-five components wearing base-ten digits. The ten-adic numbers are, in a precise sense, the two-adic numbers and the five-adic numbers side by side, and $e$ is the switch that selects one of them.

Every base has its own count #

The argument for base ten works for any base $b$: split $b$ into its prime powers, put a dial on each, and choose $0$ or $1$ on each dial.

The count is $2^{\omega(b)}$, where $\omega(b)$ is the number of different primes dividing $b$. Bases that are powers of a single prime — two, three, four, five, seven, eight, nine — have only $0$ and $1$, and their adic numbers have no zero divisors. Base six has four, like base ten: $3^2 = 9$ is written $13$ in base six and ends in $3$, and $4^2 = 16$ is written $24$ and ends in $4$. Base thirty has eight, one for each way to choose $0$ or $1$ on the dials of $2$, $3$ and $5$. The figure checks the count by brute force for every base up to 36.

Base ten is not special in having these numbers; it is special only in being the base people write in, which is why $625$ and $9376$ have been noticed by people who were never told about the Chinese remainder theorem. Numbers whose square ends in themselves are called automorphic numbers in recreational mathematics, and their supply in base ten — exactly two new ones of each length, with occasional exceptions where the new leading digit is a nought — is the dials’ four solutions with the trivial ones removed.

What the digits look like #

The two strings are defined by an equation with a two-line solution, and their digits can be computed by the million. What they look like is another matter.

Because the two strings add up to $\ldots 0001$, their digits are tied: apart from the last place, a digit of one is nine minus the digit of the other, and the figure checks that across all twenty thousand places. So the two strings have the same statistics, mirrored. Those statistics look like a fair ten-sided die’s: each digit occurs close to two thousand times, and a chi-square test of the counts gives $10.6$ on nine degrees of freedom, entirely ordinary.

That is the same observation fractions repeat and roots look random made for the digits of $\sqrt2$ and a root lifted one digit at a time made for its seven-adic digits, and it has the same status. Nothing proves that the digits are evenly spread in the long run. The string $\ldots 109376$ is the two-adic number $0$ glued to the five-adic number $1$ by the Chinese remainder theorem, and its decimal digits are what that gluing looks like when written in base ten — a mixing of a base-two expansion with a base-five one that nobody has analysed into a proof of normality.

What the dials cannot show #

The figures compute everything on dials of up to $10^{20000}$, in whole-number arithmetic, so each digit shown is exact. What they cannot show is the infinite object. A ten-adic number is the whole consistent family of last digits, and the figures show finitely many members of the family; the claims about the limit — that it is its own square, that the product is nought — are proved by the dial argument at every length, not by any computation.

The convergence they draw also needs a word of interpretation. In ten-adic arithmetic two numbers are close when they share many final digits, not when their difference is small in the ordinary sense: $5^{2^{8}}$ and the limit agree in their last ten digits, while $5^{256}$ itself is an ordinary number with 179 digits that is nowhere near anything in the ordinary sense. The iterations converge in the sense of the dials, and only in that sense.

Still open: the leading digits #

The digits of the two strings are not understood statistically, and one specific question about them has attracted attention. Each new solution of $x^2 = x$ modulo $10^n$ extends the previous one by one digit, and sometimes that digit is nought, so that the $n$-digit solution has fewer than $n$ significant digits — $0625$ is the four-digit solution, and $625$ was already the three-digit one. How often the new digit is nought, or any given digit, is the same question as the frequency of digits, and the expectation that each digit appears one time in ten, in the long run, in both strings, is supported by every computation and proved by none.

Behind it is a general question about adic numbers defined by simple equations: for which equations, and which bases, are the solutions’ digits normal? For rational solutions the digits eventually repeat, as in base ten; for these irrational ones, as for the seven-adic square root of two, the answer is believed to be “always normal” and is not known for a single case.