Download Problem of the Week #16

Problem of the Week #16
4/04/2016 to 4/17/2016
Let x1 , x2 , x3 , . . . be a sequence of nonzero real numbers satisfying
xn =
xn−2 xn−1
for n = 3, 4, 5, . . . .
2xn−2 − xn−1
Establish necessary and sufficient conditions on x1 and x2 for xn to be an integer
for infinitely many values of n.
Solution: We must choose x1 = x2 to be nonzero integers.
To begin
x2 x3
x1 x2
and x4 =
,
x3 =
2x1 − x2
2x2 − x3
and so by substitution,
x1 x2
x2
x1 x 2
2x − x2
1
=
x4 =
.
x1 x2
3x1 − 2x2
2x2 −
2x1 − x2
Repeating this process yields that
xn =
x1 x2
x1 x2
,
=
(n − 1)x1 − (n − 2)x2
n(x1 − x2 ) + (2x2 − x1 )
a fact which may be verified by induction. Since x1 and x2 are fixed, nonzero numbers,
unless x1 = x2 , |xn | will become small as n increases. That is, if x1 6= x2 then xn will tend
to 0 as n tends to infinity. So, for some subscript N , 0 < |xn | < 1 for every n > N , thus,
only a finite number of the xn ’s will be integers. Consequently, the only way to obtain an
infinite number of integer values in the sequence is to have x1 = x2 . In this case, xn = x1
for every n, so unless x1 is an integer, no xn will be an integer. Therefore, to generate an
infinite number of integers in the sequence we must choose x1 = x2 and let their common
value be any nonzero integer.
Solutions for this problem were submitted by Arghyanil Bhattacharya (Kolkata,
India), Charles Burnette (Drexel), Kipp Johnson (Beaverton, OR), Jack Kennedy
(Seattle, WA), Hari Kishan (India), Tin Lam (St. Louis, MO), Srijit Mukherjee
(Kolkata, India), and Tom O’Neil (Central Coast, CA).