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Throuuh thg decimal poiltt
Where the quadratic x2 - x has four roots
by A. B. Zhiglevich and N. N. Petrovl
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E WILL TRY TO PERSUADE YOU THAT THE
number of roots of the ecluationx2 = x is not two/
as everybody thinks, but four, as stated in the
title of this article. But first we offer the following
undisputed fact. If a natural number (in decimal notation)
ends in any of the digits 0, l, 5, or 5, its square
ends in the same digit. This also applies to the two-digit
endings 00, 01, 25, and75-thatis, if a number ends in
one of these pairs of digits, its square does too (for instance,
17 62 = 30,97 6j 2252 = 50,625 ; and so on). A similar
property holds for the three-digit endings 000, 001,
625, and376. At the same time, no other one-, two-, and
three- digit endings possess this property. What's so
special about these combinations of numbers? The
answer is given by the following theorem.
THroRru. For any natural k there are exactly four
sets of k digits (00. . .00, 00 . . .01, and two more, ending
in 5 and 6, respectively) such that if a natural number
ends in one of these sets of digits, the square of this
number ends in the same set of digits.
To eliminate any doubts, we'llgive two proofs of this
theorem. But first let's try to understand better what
we have to prove.
We're looking for an integer & 0 < x < 10&, such that
for any integer a > O the square of the number 10&a + x
(which is the general form of a number whose k-digtt
ending coincides with x) has the form 10kb + x, where b
is an integer. But (108a + xlz = tCf(tOka2 + 2axl + x2, so
our condition simply amounts to the fact that x2 - x is
divisible by 1S. Thus we've arrived at a very simple problem:
to find all integers x, 0 < x < 1#, such that x2 - x is
divisible by 1S. We must make sure that this probiem
has four solutions, and that these solutions have the properties
specified in the theorem.
First proof . The difference * - x = x(x - 1) must be
divisible by 10k : 2k5k. But the numbers x and x - 1
lSome material by V. Denisenko was incorporated in
this article with his permission.-Ed.
can't both be divisible by 2 arrd can't both be divisible
by 5. So we're left with four possibilities:
(1) x is divisible by 10ft;
(21 x - 1 is divisible by 10ft;
(3) x is divisible by 2k andx - 1 is divisible by 5k;
(4) x is divisible by 5k and x - I is divisible by 2k.
Since x < 10&, the {irst possibility can occur only i{ x = 0;
similarly, the second possibility corresponds to x = 1.
Lr case (3 ) the numberx (and in case (4) the numberx - 1 )
isoneof thenumbers a.2k, andsince x>x- 1>5k,
we know that 0 < a < 5k - 1. All these numbers yield
different remainders when dividedby 5k, because if the
numbers Zkb and2kc give the same remainders upon
division by 5k, then their difference 2k(c - b) is divisible
by 5k, and so by 10k as well, which is impossible
< 5k. But there are only 5k possible remain-
for 0 < c -b
ders, so each of them is represented once among our
numbers 2ka. This means that exactly one of these
numbers-say,Zkar-'has the remainder l, so that for
x = Zkav the numberx - I is divisible by 5ft. Similarly,
exactly one of these numbers-say, Zkar-has the remainder
5k - 1 when divided by 5k, and-so, for x - I =
2ka, the number x is divisible by 5k.
This conclusively shows that in each of our four
cases there is one and only one x satisfying our conditions,
so the problem has exactly four solutions. It's
also clear that in case (3) the number x ends in 6, and
in case (4) it ends in 5. The proof is complete.
Second proof . We'll take it for ganted that the problem
has no more than four solutions. (We know this from
the first proof, but it can be proved directly as we11.)Also,
we'll use induction over k: we'll assume that we've already
found the numbers xu_, and )27._ 1
ending in 5 and
6, respectively, such that 0 (xo_, . lG- I, 0 ( /r_ r
. 1S-',
and the differences x? _ t - xx _ , and yoz_ f yx _, are divisible
by lgr- 1 (we create a basis for the inductionthat
is, the existence of x, and yr- by setting xr = 5,
0U[]tlIUil/rtlIltRt 17