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 ^{n-1}^{n}^{n}^{4k ± 1}In 1930 the writer proved, as a by-product of an extended theory of Lucas's functions U and
_{n}V, that the test which Lucas
gave for _{n}^{4k + 1}^{4k - 1}
THEOREM A.
S_{1} = 4,
S_{2} = 14,
S_{3} = 194,
...,
S, ...,_{k}
S =
_{k}S_{k}^{2}_{- 1}-2In 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.
We define two sequences of integers U and
_{r}V by_{r}
U =
(_{r}a) / (^{r} -
b^{r}a - b),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
LEMMA 1.
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.
Proof.
To prove (7) we expand U
as follows:_{p}
All the binomial coefficients are divisible by
U
3_{p}^{½ (p - 1)}
(3/p) (mod p).
V
in like manner, thus_{p}
In this case all the binomial coefficients except the first are divisible
by
LEMMA 3.
U_{p + 1} =
2U,_{p} +
V_{p}
-4
U_{p + 1}
U_{p - 1} =
4U_{p}^{2} -
V_{p}^{2}
4(±1)^{2} -
40
(mod p).
_{k}, ...
_{k} =
2^{2k - 1}
S. Then it
is sufficient to show that
_{k}_{n - 1}
is divisible by N. Since
S_{k + 1} =
S_{k}^{2} - 2,
_{k + 1} =
_{k}^{2} -
2^{2k + 1}.
r = 2^{k} in (5) we get
V_{2k + 1} =
(V_{2k})^{2} -
2^{2k + 1}.
V_{2} = 8 =
_{1}. Hence, in general,
_{k} =
V_{2}.
V_{2n - 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)}.
since 2 is a quadratic residue of N = 8x - 1. But (9)
follows from Lemma 2 and (3). In fact
V_{N + 1} =
V_{N} V_{1} +
12U_{N} U_{1} =
2V_{N} +
12U_{N}.
p = N, we note that
N 1 (mod 3)
and that
N) = -(N/3) = -(1/3) = -1.
V_{N + 1} =
V +
6_{N}U
2 - 6
-4 (mod _{N}N).
_{n - 1} =
2^{2n - 2}
S_{n - 1} =
V_{2n - 1}.
p be any prime factor of N and let
be the rank of apparition of p. Then p divides
U_{2n} since N divides
U_{2n - 1}
V_{2n - 1}, which is
U_{2n}
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_{2n - 1} as well as
V_{2n - 1}. This is impossible by
(6) since p is odd. Hence =
2^{n}.
By Lemma 3,
p > - 1 =
2^{n} - 1 = N.
p = N, so that N is a prime.
In his paper Dr. Western gives a list of Mersenne numbers
2
n = 157, 167, 193, 199, 227, and 229.
Lehigh University,
¹ Received 5 November, 1934; read 15 November, 1934. |

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