Internat. J. Math. and Math. Sci.
Vol. i0 No. 2 (1987) 409-412
409
EVEN PERFECT NUMBERS AND THEIR EULER’S FUNCTION
SYED ASADULLA
Department of Mathematics and Computin Sciences
St. Francis Xavier University
Antigonish, Noga Scotia.
(Received February 21, 1985, and
ABSTRACT.
B2G 1C0.
Canada.
in revised form,
June
23,
1986)
The purpose of this article is to prove some results on even perfect
numbers and on their Euler’s function.
The results obtained are all straightfor-
ward deductions from well-known elementary number theory.
KEY WORDS AND PHRASES.
Perfect number; triangular number- Euler’s function; number
of divisors function.
1980 MATHEMATICS
SUBJECT CLASSIFICATION CODE.
INTRODUCTION.
A positive integer
1.
IOAdO
is called a perfect number if it is equal to the sum of its
positive divisors excluding itself.
The n
th
triangular number is the sum of the first n-positive integers
I
k
F
.
n
k=l
Euler’s function
to
@(n)
is the number of positive integers less than or equal
and relatively prime to
n
T(n).
n(n+l)
n.
The number of divisors function
is the number of positive divisors of
d(n)
MAIN RESULTS.
The proof of the following Theorem
can be found in many elementary number
theory books; see, for example, [l:p. 98].
THEOREM 1.
If n is an even perfect number, there exists a prime
that
THEOREM 2.
Pk
If
T(pl)
is any even perfect number, where
is the first prime in the sequence
then
2P-1
such
p-1(2p-1).
2
n
T(Pk)
{P2,
P3
Pj
Pl
is prime, and if
} where pj
2Pj_1+1,
is the next even perfect number.
It follows from Theorem 1 that an even perfect number is of the form
can be written as T(pI),
2n-I is prime. Now,
=2n-1. Let Pi be any composite term of the sequence
PROOF.
2n-l(2n-l),
where
pl
.....
Pi
It can be shown that
Pj
2n-l(2n-l)
where
2Pj_1+1.
n+i-1 -1,
2
Now, it follows from Theorem
using the facts
that
T(Pi)
Pi
2n-1’
n+i-1
2n+i-2(2
and
-I)
is
410
S. ASADULLA
Let
not an even perfect number.
As before,
}.
pj
2m-1(2m-l),
is of the form
2n+k-I
Pk
T(Pk) 2n+k-2(2
thus T(Pk) is an even
n+k-I
Observe that
-1.
2m-1
where
(P2,
be the first prime in the sequence
Pk
is prime and
P3,
-1)
perfect number by Theorem 1.
EXAMPLE.
TC3
C3)Cd)
T(31)
THEOREm3.
C31)(32)
If
n=
then,
2 (m+1)/2
2k,
+ (2k-1)
2 (m-1)/2
(127)(128)
+ 3
1
n
+
+
+ 3
+ 2
[2 + 4 +
2o(1 o + 2 +
k(2k+l)
2k-(k+l)
+
[2 (m+l)/2 -1] 3.
Now, consider
=[
(2k)(2k+l)]2
(2k-1)
+
+ C2k) ]
+ k)
+ k)
8(I + 2 +
S[ k(k+l)]Z
k(2k
1).
follows that 1+ 3 +...+ [2 {m+1)12
it
8128
2 {m-1)/2.
k
where
k-C2k+l)
k(2k+l)
k-(2k+l)
kZ(2k+l) z
2
28.
[1+2+3+...+(2k-1) + (2k)] 2
+ (2k)
which implies that
k
T(127)
496
2m-l(2m-l),
Observe that
Since
C7)C 8)
TO7
6,
2m-1(2m-l)--
1] 3
The following Corollary I, follows from Theorem 3.
COROLLARY 1.
If
n
n
EXAMPLE.
is an even perfect number
+ 3
1
+ 3
496
+
+
C2
(p+I)/2
+ 7; p
+ 5
2P-1(2P-1),
1]
.
then
5.
The proof of the followingTheorem 4 can also be found in many elementary
1:
number theory books; see, for example
THEOREM 4.
If
n
Pt
51
63.
p.
Pk%
P2a
1.
1
P
P’ P
are distinct primes and
Pk
-
a
a, a2
are positive integers.
k
As a consequence of Theorem 4, one can easily obtain Theorem 5, Corollary 2,
and Corollary 3
2P-1(2p
#(.) P-(2
THEOREM 5.
n
COROLLARY 2.
EXAtLE.
It n
THEORE 6.
If
n
If
then
PROOF.
@(nt
pi-1
@C 2
n,
n...,
pi.
4032
8128
n
@(n)
n
4p-1.
@(n)
n
2p-2.
46
is an even perfect number, then
are k-distinct even perfect numbers,
n
k
nk)= .k-1
@(nk).
@(nt) @(n)
nk)
p.-1
(2p -1) 2
+
@E2
PI+P- +
@(2
P-
,h,
is an even perfect number, then
@(nl n2
PI+P- +
).
@(26 @(127)
@{S128)
COROLLARY 3.
is an even perfect number if and only if
1)
+
Pk-k
Pk-k
Pk-1
2p- -1)
2
P
Pl
-I) (2
(2
Pl
@(2
Pk -I)]
Pk
@(2 -1)
(2
-1)
P-
-1) @(2
pk
(2 -1)]
-I)
n
EVEN PERFECT NUMBERS AND THEIR EULER’S FUNCTION
+
Px+P+
-k-1
Pk
=2
p
p
-2) (2
(2
(2Pk -2)
-2)
p-I
p-I
pi-1
2
{2
-1)
k-1
2
p-I
{2
2
411
Pk-1
-1)
2
2Pk-1
-1)
k-1
@(nl) @On2)
9.
@(nk).
The following Theorem 7 is proved in many books on elementary number theory;
see, for example,
[1:
p.
If
n
9].
k
THEORE 7.
a
R
i=1
k
are distinct primes and
k
i
a
R (I +
d{n)
then
Pi
i=1
Pi’
i=l
d(n)
are positive integers, and
k
i=1
i.
where
ai),
is
the number of divisors function.
k
THORE 8.
If
a
R
n
i=1
2p-I(2p
i
and
Pi
d{n}
is an even perfect number
I), then
i}
p
ii}
a
2
uj
o.
a
2
iil)
k.
j
I
k
iv)
for exactly one
-I,
I>
where
O,
k
ff
i=l
(1 +
aj).
J
k,
Thus
(1 +
and
(2p -1)
p
k
{1
# J,
/
I >
2p-l,
t)
J
X
O;
k
{1 +
a
p-1
2
(2p -1),
i)
which tmplie that
0
gj
1
(l+ai),
and
i=1
/
and
j
which
say
such that
that is,
U
2
i
p-I
i
k,
i
X for some X and exactly one
(2 -1)
aj)
i=l
X
k
p-1
2
(2p -1),
(l+ai)
(2p -1) divides exactly one of the factors
implies that
< J
J.
I
k,
1
From Theorem 5, one obtains d(n)
PROOF.
1
such that
1.
p
Ui
i=1
{2p -I} -I
which is
k
Observe that
p-1
k-1
or
p
for exactly one
(2p
1)
1
,
k-1
i=1
i
k,
which is
such that
1
for exactly one
J
since
i >
(t).
Now,
{1 +
aj)
{2p -1)
J
k and
j
0
implies that
such that
0
1
for
i
k
J
and
O.
and
O,
aj
Thus,
2J
which proves
412
S.
Finally,
a
=2
1,
1
ACKNOWI.
+ a
2
i
k,
1
i
<_
J,
i
ASADULLA
k,
i >
i
O,
J,
i >
0
implies that
which proves (tit).
The author wishes to tl-tk the referee for his many helpful
sugge s t ions.
I.
SHOCKL, J.E.,
York, N.Y.
REFERENCES
Introduction to Number Theory, Holt, Rinehart and Winston, New
1967.