Vol. 11 No 1 - Pi Mu Epsilon

20 GARZA AND SHINKEL
TIME Journal, Vol. 11, No. 1, pp 21- 22, 1999. 21
[3] K. KURATOWSKI, Sur le probleme des courbes gauches en topologie, Fund. Math. 15 (1930)
270-283.
[4] J. MYCIELSKI, Sur le coloriage des graphes, Colloq. Math. 3 (1955) 161-162.
Gina Garza, Department of Mathematics and Statistics, Western Michigan University, Kalamazoo,
MI 49008-5152.
Natascha Shinkel, Pearce Hall 214, Central Michigan University Mount Pleasant, MI 48859.
natascha.shinkel@cmich.edu
Gina Garza is now a senior at Western Michigan University majoring in mathematics
and secondary education. She also coaches junior varsity tennis at a nearby
high school.
N atascha Shinkel worked on this paper while a senior mathematics major at
Western Michigan University majoring in mathematics with a minor in dance. She is
now a graduate student at Central Michigan University.
@Herman Servatius, 1996.
Have you ever wondered
"What is Art?" A theory
of neuro-scientists which has
been seized enthusiasticly by
popular culture is that the
left hemisphere of the brain
is the "mathematical side" ,
governing logic, counting and
sorting, while the right hemisphere
is the "artistic side" ,
governing images and creativity.
Where in this taxonomy
can we place the "elegant
mathematical proof" or
the "beautiful" mathematical
figure?
Has the image at the left side
of this page come from the
right side of the brain? Is it
art? It is titled "Mathematics
Kitch" and is an Escher­
Fractal 3D Stereogram, e.g. ,
a 3D stereogram of the Mandelbrot
set based on a tesselation
by M.C. Escher.
The liME Journal invites those of you who paint, draw, compose, or otherwise use
the other side of your brains to submit your mathematically inspired compositions.
WHEN ARE AN - 1 AND AM- 1 AMICABLE?
FLORIAN LUCA*
For any positive integer k let a(k) be the divisor sum of k. Two positive integers
s and t are called amicable if a(s) = a(t) = s + t. A positive integer s which is
self-amicable is called perfect.
In [3], we showed that no Fibonacci or Lucas number is perfect. In [4], we showed
that no Fermat number, i.e. a number of the form 2 2 n + 1 for some integer n 2: 0,
can be either perfect or part of an amicable pair.
Let a> 1 be a positive integer. In this note, we prove the following:
THEOREM 1. If am- 1 and an- 1 are amicable, then m = n = 1 and a- 1 is
perfect.
Let us remark first that:
LEMMA 2. If s > 1 is a positive integer, then a(s) ::; 3s 2 /4.
Proof Since a(st) ::; a(s)a(t) for any two positive integers s and t, it suffices to
check the asserted inequality only when s is prime. But if s is prime, then a(s) =
s + 1 ::; 3s 2 /4 with equality if and only if s = 2. D
Proof of the Theorem. Since the involved expressions are symmetric in m and n ,
we shall always assume that m ::; n.
Assume first that m = 1. We need to show that n = 1.
Assume that n > 1. Suppose first that a= 2. Since an -1 = 2n -1 and a - 1 = 1
are amicable, we get a(2n - 1) = a(l) = a(l) = 1 which forces 2n - 1 = 1, hence
n = 1.
Suppose now that a> 2. Since an-1 and a-1 are amicable, we get a(a- 1) = an+
a-2. Since by the Lemma, a(a - 1)::; i(a - 1)2 , it follows that an+a - 2::; i(a - 1) 2 •
Since n 2: 2, we get a 2 +a- 2 ~ i(a -1) 2 , which is equivalent to a 2 +lOa::; ll,which
is certainly impossible for a> 2.
From now on, we assume n > 1. Suppose first that n = m. In this case, an - 1
is perfect. It follows easily that a is odd. Indeed, if a is even, then an - 1 is an
odd perfect number which is congruent to -1 modulo 4. On the other hand, it is
well-known that if an odd perfect number exists, then it should be of the form px 2
for some prime p = 1 (mod 4). Hence, a is odd. In this case, an - 1 is an even perfect
number, therefore an - 1 = 2P-l (2P- 1), where 2P- 1 is a Fermat prime. Write
Since
an - 1 a 2 - 1
--- > -- = a+l >a-1,
a-1 - a-1
it follows easily that the prime 2P -1 divides (an -1)/(a-1). Hence (an -1)/(a-1) =
2u(2P - l)and a - 1 = 2v where u+v = p-1. When n is odd, the number (an-1)/(a- 1)
is odd as well, hence u = 0. It follows that a= 2P-l + 1, and
an -1 a 3 - 1
2P- 1 = -- 2: --= a 2 +a+ 1 = (2P-l + 1) 2 + (2p-l + 1) + 1,
a - 1 a - 1
*Mathematical Institute, Czech Academy of Sciences