nL;

.....00000
.....00132
. . . ..99868
Figure 3 Figure 4
ecluation x2 = x has exactly four solution in supernumbers:
0, 1, X andY.
And that's the result we've been waiting for.
llow is fialRossible?
If the operations on supernumbers obey the usual
rules, the usual proof of the f.act that the equation
* - x = 0 has two solutions must work equally well
with supernumbers. Let's see if they really do. As before,
* - x = x(x - l)-there's nothing to rule out this
f.actorization. Therefore ,ilf.* -x:
......052s
x......9376
......3750
......437s
...... 1875
.....5525
0000
0, then eitherx = 0 or
x- 1 = 0. But why? How do we know that the product
ab of supernumbers a andb is zero only if one of them
is zero? In (act, that's wrong! We know perfectly well
that XY: 0!
Indeed, xo is divisible by 5k and yo is divisible by 2k,
so the productxT,)zp is divisible by lOk- that is, it ends
in k zeros. So multiplying X and Y digit by digit, we'll
get only zeros in the result. To dissolve any possible
doubts you may have, look at frgure 4.
If a+0,b+O,butab:0, thena andb arecalledzero
divisors. There are no zero divisors among ordinary
numbers, but we can find them among supernumbers-and
that's the gist of the matter.
By the wa, now we can verify once again that X andY
satisfy our equation: since X + Y = 1 (rememb et, xp+ /p:
10ft + 1- see exercise l), X(X - 1 ) = -X( 1 - Xl : -XY : O,
and, similarly, -Y(Y - 1) : 0.
The equation P=I
Ifx2 :x, then# :* - x : X. x =* = x, * : * . X : X
. x: *: & andso on: x5 = x, * = x,... . So ourfour
solutions to * : x satisfy xm = x for arry m. But does
the equation xm = x have any other roots?
Let's startwith m = 3. Besides 0, I, X, and I the
equation # = x has a solution X - Y:
lx - Yl3 : vz - Y - \xY(x - Yl = x - Y,
because XY = O, alternatively, we could prove that
(X -Ylz = l, which also implies lX -Yl3 = X -Y.In addition,
we can reverse the sign of any root of the equation
# : x and this will yield four other solutions: -1,
-X, -Y, Y - X. A11 in all, we've found nine solutions:
0, 1, -1, X, -X, Y, -Y, X - Y, Y - X.
Exercise 2. Prove that the equation f = x has no
other solutions.
The case m: 4 (and in general the case of any even
m-see the theorem below) is not as interesting.
Exercise 3. Prove that the equations * = x andl2:
x have the same solutions.
In the case m = 5 new solutions emerge. To describe
them, it will suffice to display one of them. Consider a
sequence zo in which 1= 2 and any term zo consists of
the last k digits of zf _ r:
zt : 2, zr: 32, z, : 432, ... .
Exercise 4. Prove that zo is obtained from zo_rby
adding one digit on the left, and that zf - zo is divisible
by 10ft for all k.
The numbers zo define the supernumber
z = ...9879t85432.
Exercise 5. The equationxs -xhas fifteen solutions:
the nine solutions of # - x (see exercise 2) and the
supernumbers Z, -2, X - Z, Z - X, X + Z, -X - Z.
For greater values of m new solutions don't appear.
The reader may try to prove the following theorem.
THronil,{. If m is even, the equation xm : x has the
same solutions (in supernumbers) as x2 = x. If m has
the form 4n - 1 , this equation has the same solutions
as # - x. And if m: 4n + 1, then xm : x has the same
solutionsasx5-x.
Here is one more statement conceming the equation
x- : l:If m is odd, thenx- : t has only one solutionx: 1.
If m is even but not divisibleby 4, then xm = I has four
solutions: l, -1, X - Y, Y - X. If m is a multiple of four,
then xm = I has eight solutions: 1, -1, X - Y, Y - X, X - Z,
Z-X,X+2,-X-2.
fuo mol,e eqtlatiolr$
Exercise 6. For what (ordinary) numbers m does the
equation mx : I have a solution in supernumbers?
(That is, what ordinary fractions can be found among
supernumbers?)
Exercise 7. Is the statement of Fermat's Last Theorerrl-"If
n is an (ordinary) integer greater than 2, then
the equatiort xn + yn : zn has no roots"-true for
supernumbers x, y, z7
Conclusion
If the notion of supernumbers rang a bell for you,
don't be surprised. They are known to mathematicians
as "integer lO-adic numbers." A11 the troubles-or, to
put it more positively, surprises-that we've encountered
are due to the fact that 10 is a composite number.
If we had consideredp-adic numbers with a prime
p (which are defined the same way except that we'd
have to write them using the number system with the
base p), our equation * = x would have had two solutions,
as it should. Actually, if our number system had
a prime base, we wouldn't have had enough material
foi an article like this.
O
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