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D. H. L EHMER ¹. An examination of the literature on even perfect numbers since Euclid's time reveals only two contributions to the subject which are of genuine importance. The first of these is the celebrated theorem of Euler to the effect that every even perfect number is of Euclid's type In 1930 the writer proved, as a by-product of an extended theory of Lucas's functions U n and V n , that the test which Lucas gave for
THEOREM A.
The number
N =
In 1932 Dr. A. E. Western ² gave a proof of this theorem by means of the theory of algebraic numbers. It is the purpose of this paper to give a very elementary yet self-contained proof of Theorem A. Notation and definitions. Let
We define two sequences of integers U r and V r by
V r = a r + b r .
DEFINITION. The rank of apparition of the odd prime p is the least positive subscript
(if it exists) for which
U is divisible by p.
In what follows p always represents a prime greater than 3. Before proving Theorem A it is convenient to establish three lemmas.
LEMMA 1.
If
Proof.
Let S be the set of all subscripts r for
which p divides U
r
.
From (1) and (2) it follows that if r and s are members
of S, so also are r ± s.
Hence S coincides with the set of all integer multiples of its
least positive member LEMMA 2.
Proof. To prove (7) we expand U p as follows:
![]() All the binomial coefficients are divisible by p, except the last which is equal to unity. Hence
3
½ (p - 1)
(3/p) (mod p).
![]() In this case all the binomial coefficients except the first are divisible by p. Hence (8) follows at once.
LEMMA 3.
If Proof. It is obviously sufficient to prove that p divides U p + 1 U p - 1 . From (1) and (2) with r = p and s = 1 we have, since U 1 = 1 and V 1 = 2,
-4U p - 1 = 2U p - V p .
4(±1)
2
-
4 0
(mod p).
Proof of Theorem A. We are now in a position to prove Theorem A without difficultly. The proof is in two parts. Proof of necessity. Let N = 2 n -1 be a prime. We have to show that S n - 1 is divisible by N. Instead of the series S k we may consider the series
k, ...
k
=
2
2
k - 1
S
k
. Then it
is sufficient to show that
n - 1
is divisible by N. Since
k + 1
=
k
2
-
2
2
k
+ 1
.
1
. Hence, in general,
k
=
V
2
k
.
since 2 is a quadratic residue of N = 8x - 1. But (9) follows from Lemma 2 and (3). In fact
1 (mod 3)
and that
2 - 6
-4 (mod N).
Proof of Sufficiency. Let S n - 1 be divisible by N = 2 n - 1. Then N divides
n - 1
=
2
2
n - 2
S
n - 1
=
V
2
n - 1
.
be the rank of apparition of p. Then p divides
U
2
n
since N divides
U
2
n - 1
V
2
n - 1
, which is
U
2
n
by (4). By Lemma 1, divides
2
n
. On the other hand,
does not divide
2
n - 1
, for
otherwise, by Lemma 1, p would divide
U
2
n - 1
as well as
V
2
n - 1
. This is impossible by
(6) since p is odd. Hence =
2
n
.
By Lemma 3,
- 1 =
2
n
- 1 = N.
In his paper Dr. Western gives a list of Mersenne numbers 2 n - 1 examined by Lucas's tests. Since this list was made, Mr. R. E. Powers ³ has applied Theorem A to 2 241 - 1 and finds this number to be composite. This leaves the characters of 2 n - 1 (for n < 257) undetermined for
Lehigh University,
¹ Received 5 November, 1934; read 15 November, 1934. |