ON LUCAS'S TEST FOR THE PRIMALITY OF MERSENNE'S NUMBERS
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
2
n-1
(2
n
-1),
where 2
n
-1 is
a prime. The second contribution is that of Lucas who gave a pair
of theorems governing the primality of
2
4k ± 1
-1.
A great deal of confusion exists in Lucas's writings about the exact
enunciation and actual proofs of these tests for primality.
Nevertheless, it is evident that Lucas was in possession of the facts
needed to prove the sufficiency of his tests. The confusion arose
from the fact that he was unable or unwilling to consider the
necessity also.
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 2
4k + 1
-1
is applicable also to
2
4k - 1
-1, and that
his condition for primality is both necessary and sufficient. In fact
the theorem may be stated as follows.
THEOREM A.
The number
N =
2
n
-1,
where n is an odd prime, is a prime if ,and only if, N divides the
(n-1)-st term of the series,
S
1
= 4,
S
2
= 14,
S
3
= 194,
...,
S
k
, ...,
where S
k
=
S
k
2
- 1
-2
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
U
r
=
(a
r
-
b
r
) / (a - b),
V
r
=
a
r
+ b
r
.
The following identities can be easily verified by using the above
definitions.
(1) |
2U
r + s
=
U
r
V
s
+
V
r
U
s
.
|
(2) |
(-2)
s + 1
U
r - s
=
U
s
V
r
-
U
r
V
s
.
|
(3) |
2V
r + s
=
V
r
V
s
+
12U
r
U
s
.
|
(4) |
U
2r
=
U
r
V
r
.
|
(5) |
V
2r
=
V
r
2
+
(-2)
r + 1
.
|
(6) |
V
r
2
-
12U
r
2
=
(-2)
r + 2
.
|
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 is the rank of
apparition of p, then p divides U
r
if, and only if, divides r.
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 . This proves Lemma 1.
LEMMA 2.
(7) |
U
p
(3/p) (mod p).
|
(8) |
V
p
2 (mod p).
|
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
U
p
3
½ (p - 1)
(3/p) (mod p).
To prove (8) we expand V
p
in like manner, thus
In this case all the binomial coefficients except the first are divisible
by p. Hence (8) follows at once.
LEMMA 3.
If is the rank of apparition of p, then
< p + 1.
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,
2U
p + 1
=
2U
p
+
V
p
,
-4U
p - 1
=
2U
p
-
V
p
.
Using Lemma 2, we have
-8U
p + 1
U
p - 1
=
4U
p
2
-
V
p
2
4(±1)
2
-
4 0
(mod p).
Hence Lemma 3 is proved.
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
8, 56, 3104, ..., k, ...
in which
k
=
2
2
k - 1
S
k
. Then it
is sufficient to show that
n - 1
is divisible by N. Since
S
k + 1
=
S
k
2
- 2,
we have
k + 1
=
k
2
-
2
2
k
+ 1
.
Now if we put r = 2
k
in (5) we get
V
2
k + 1
=
(V
2
k
)
2
-
2
2
k
+ 1
.
Moreover V
2
= 8 =
1
. Hence, in general,
k
=
V
2
k
.
We have to show then that V
2
n - 1
=
V
½ (N + 1)
is divisible by N.
But from (5), with r = ½ (N + 1), we have
V
N + 1
=
V
½
2
(N + 1)
-
4 · 2
½ (N - 1)
.
Hence it is sufficient to show that
(9) |
V
N + 1
-4 (mod N),
|
since 2 is a quadratic residue of N = 8x - 1. But (9)
follows from Lemma 2 and (3). In fact
2V
N + 1
=
V
N
V
1
+
12U
N
U
1
=
2V
N
+
12U
N
.
To apply Lemma 2 with p = N, we note that
N 1 (mod 3)
and that
(3/N) = -(N/3) = -(1/3) = -1.
Hence we have
V
N + 1
=
V
N
+
6U
N
2 - 6
-4 (mod N).
Hence (9) is established. This completes the proof of necessity.
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
.
Now let p be any prime factor of N and let
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,
p > - 1 =
2
n
- 1 = N.
Hence p = N, so that N is a prime.
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
n = 157, 167, 193, 199, 227, and 229.
Lehigh University,
Bethlehem, Pa., U.S.A.
¹ Received 5 November, 1934; read 15 November, 1934.
² "On Lucas's and Pepin's tests for the primeness of Mersenne
numbers",
Journal London Math. Soc., 7 (1932), 130-137.
This paper contains a list of references to papers on the subject.
³
Bull. Amer. Math. So., 40 (1934), 883.
|