Download Proof that an Infinite Number of Mersenne Prime

Proof that an Infinite Number of Mersenne Prime Numbers Exit
Stephen Marshall
29 Aug 2014
Abstract
This paper presents a complete and exhaustive proof of the infinitude of Mersenne
prime numbers. The approach to this proof uses same logic that Euclid used to prove
there are an infinite number of prime numbers. Then we prove that if p > 1 and d > 0
are integers, that p and p + d are both primes if and only if for integer n (see
reference 1 and 2):
1 (−1)dd!
1
1
𝑛 = (𝑝 − 1)! � +
�+ +
𝑝
p + d
p p+d
( + )
( )
We use this proof for d = 2𝑝 𝑘 𝑚 - 2𝑝 𝑘 to prove the infinitude of Mersenne prime
numbers.
1
Introduction
Marin Mersenne, (8 September 1588 – 1 September 1648) was a French theologian,
philosopher, mathematician and music theorist, often referred to as the "father of
acoustics" (Bohn 1988:225). Mersenne was "the center of the world of science and
mathematics during the first half of the 1600s." The Mersenne primes are named after
the Mersenne who first studied them in the early 17th century.
In Number Theory, a Mersenne prime is a prime number of the form
.
This is to say that it is a prime number which is one less than a power of two. The first
four Mersenne primes are 3, 7, 31 and 127.
If n is a composite number then so is 2n − 1. The definition is therefore unchanged when
written
where p is assumed prime.
More generally, numbers of the form
without the primality requirement
are called Mersenne numbers. Mersenne numbers are sometimes defined to have the
additional requirement that n be prime, equivalently that they be pernicious Mersenne
numbers, namely those pernicious numbers whose binary representation contains no
zeros. The smallest composite pernicious Mersenne number is 211 − 1.
As of May 2014, 48 Mersenne primes are known. The largest known prime number
(257,885,161 − 1) is a Mersenne prime. Since 1997, all newly found Mersenne primes have
been discovered by the “Great Internet Mersenne Prime Search” (GIMPS), a distributed
computing project on the Internet.
Proof:
We shall select the set of Mersenne primes and make an assumption that this set of
Mersenne prime numbers are finite. We shall then prove that this assumption is
incorrect and by contradiction prove that an infinite number of Mersenne primes exist.
2
First, according to our finite set of k Mersenne primes, we shall let p be equal to the last
Mersenne prime in the finite set, which is shown below:
( )
Let, p = 𝑀𝑝(𝑘) = 2𝑝 𝑘 – 1
Where 𝑀𝑝(𝑘) is the kth Mersenne prime in the finite set.
( + )
Let, 𝑀𝑝(𝑘 + 𝑚) = 2𝑝 𝑘 𝑚 – 1 , where m is an integer.
𝑘 + 𝑚)
And let, 𝑀𝑝(𝑘 + 𝑚) = 2𝑝(
– 1, where, 𝑀𝑝(𝑘 + 𝑚), is a Mersenne number not in
the finite set of Mersenne primes, therefore, 𝑀𝑝(𝑘 + 𝑚) must be composite and not
prime.
( )
( + )
Then we shall let p + d = 2𝑝 𝑘 – 1 + d = 2𝑝 𝑘 𝑚 – 1 = 𝑀𝑝(𝑘 + 𝑚)
Where 𝑀𝑝(𝑘 + 𝑚) is a composite Mersenne number by definition and d is a positive
integer.
( + )
( )
Then for our case, d = 2𝑝 𝑘 𝑚 - 2𝑝 𝑘
We will prove that p and p+ d are both primes if and only if for integer n
1 (−1)dd!
1
1
𝑛 = (𝑝 − 1)! � +
�+ +
𝑝
p + d
p p+d
3
( )
( + )
Therefore, we must show that p + d = 2𝑝 𝑘 – 1 + d = 2𝑝 𝑘 𝑚 – 1 = 𝑀𝑝(𝑘 + 𝑚) is prime to
( + )
prove that 2𝑝 𝑘 𝑚 – 1 = 𝑀𝑝(𝑘 + 𝑚) is prime, this will prove that 𝑀𝑝(𝑘 + 𝑚) is prime instead
of composite which will show there is a Mersenne prime that is outside our assumed set of finite
Mersenne primes.
We shall use Euclid's logic that he used to prove there are an infinite number of prime
numbers to prove there are an infinite number Mersenne primes.
First we shall assume there are only a finite number of k Mersenne primes for all positive
integers, specifically the last Mersenne prime in our finite set is;
𝑀𝑝(𝑘) = 2𝑝(𝑘) – 1
( )
( )
( )
( − )
( )
1) Say, 2𝑝 1 – 1, 2𝑝 2 – 1, 2𝑝 3 – 1, . . . . , 2𝑝 𝑘 1 – 1, 2𝑝 𝑘 – 1
( )
( )
( )
( − )
( )
2) Let N = (2𝑝 1 – 1)(2𝑝 2 – 1)(2𝑝 3 – 1), . . . . (2𝑝 𝑘 1 – 1)(2𝑝 𝑘 – 1) +1
By the fundamental theorem of arithmetic, N is divisible by some prime q. Since N is the
product of all existing Mersenne primes plus 1, then this prime q cannot be among the k
()
2𝑝 𝑖 – 1
that make up the k Mersenne primes since by assumption these are all the Mersenne
primes that exist and N is not divisible by any of the k
()
2𝑝 𝑖 – 1 Mersenne primes.
N is clearly seen not to be divisible by any of the k Mersenne primes. First we know that
2 is a prime number that is not in the set of finite Mersenne primes since the first
Mersenne prime is 3. We also know that 2 is the only even prime number, therefore,
the finite set of k Mersenne primes are all odd numbers. Since the product of odd
numbers is always odd, then the product of all the k in our Mersenne primes in the finite
4
set of Mersenne primes is an odd number. Since N is product of all the k Mersenne
primes + 1, then N is an even number, and since all the Mersenne primes are odd
numbers and N is even, then N is not divisible by any of the k Mersenne primes.
Therefore, q must be another prime number that does not exist in the finite set of
Mersenne prime numbers. Therefore, since this proof could be repeated an infinite
number of times we have proven that an infinite number of prime numbers q exist
outside of our finite set of Mersenne primes.
Now we must prove that at least one of these infinite prime numbers, q, is a Mersenne
( )
prime. We will pick a prime number p = 𝑀𝑝(𝑘) = 2𝑝 𝑘 – 1 which is the last Mersenne
prime from our finite set of Mersenne primes and we will need to prove that there exists
a prime 2𝑝(
𝑘 + 𝑚)
– 1, that is also a Mersenne prime. Both p and p +2k do not exist in
the finite set of Mersenne primes. Note we are not proving this for all q primes outside
the finite set of Mersenne primes, we are only picking one number 2𝑝(
𝑘 + 𝑚)
– 1 from
the infinite set of primes outside our finite set and then we shall prove that 2𝑝(
𝑘 + 𝑚)
–1
( )
is prime given that 2𝑝 𝑘 – 1 is Mersenne prime, this will show that at least one
Mersenne prime exists outside our finite set of Mersenne primes.
Now we shall proceed to prove 2𝑝(
𝑘 + 𝑚)
– 1 is prime as follows:
We will prove that if p > 1 and d > 0 are integers, that p and p + d are both primes if
and only if for positive integer n (see reference 1 and 2):
1 (−1)dd!
1
1
(𝑝
𝑛 = − 1)! � +
�+ +
𝑝
p + d
p p+d
5
Proof:
The equation above can be reduced and re-written as:
3)
(𝑝 – 1)! + 1 (−1)𝑑𝑑! (𝑝 − 1)! + 1
+
p
p+d
Since (p + d − 1)! = (p + d − 1)(p + d − 2) · · · (p + d − d)(p − 1)!, we have
(p + d − 1)! ≡ (−1)d d!(p − 1)! (mod p + d), and it follows that equation
4 above is an integer if and only if:
4)
is an integer.
(𝑝 – 1)! + 1 (p + 𝑑 – 1)! + 1
+
p
p+d
From Wilson’s Theorem, if p and p + d are two prime numbers,
then each of the terms of, equation 4 above, is an integer, which proves the
necessary condition. Wilson’s Theorem states:
That a natural number n > 1 is a prime number if and only if
.
That is, it asserts that the factorial
is one less than a multiple of n exactly when n is a prime number. Another way of
stating it is for a natural number n > 1 is a prime number if and only if:
When (n − 1)! is divided by n, the remainder minus 1 is divides evenly into (n-1)!
Conversely, assume equation 4 above, is an integer. If p or p + d is not a prime,
then by Wilson’s Theorem, at least one of the terms of (4) is not an integer. This
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implies that none of the terms of equation 4 is an integer or equivalently neither of p
and p + d is prime. It follows that both fractions of (4) are in reduced form.
It is easy to see that if a/b and a′ /b′ are reduced fractions such that
a/b + a′ /b′ = (ab′ + a′ b)/(bb′ ) is an integer, then b|b′ and b′ |b.
Applying this result to equation 4 , we obtain that (p + d)|p, which is impossible. We
may therefore conclude that if equation 4 is an integer, then both p and p + d must
be prime numbers. Therefore, the equation below is proven since it can be
reduced to equation 3 above.
Therefore, since the below equation can be reduced to equation 3 above, we have
proven that if p > 1 and d > 0 are integers, then p and p + d are both primes if and
only if for positive integer n:
1 (−1)dd!
1
1
𝑛 = (𝑝 − 1)! � +
�+ +
𝑝
p + d
p p+d
( )
( + )
For our case, p + d = 2𝑝 𝑘 – 1 + d = 2𝑝 𝑘 𝑚 – 1
( )
Therefore, p = 2𝑝 𝑘 – 1 = 𝑀𝑝(𝑘) is the kth Mersenne prime in the finite set of Mersenne
primes. Therefore p is prime.
( + )
( )
Then d = 2𝑝 𝑘 𝑚 - 2𝑝 𝑘
Therefore, we must prove that n is an integer for integer’s k and m, of which k denotes
the last Mersenne prime in our assumed finite set of Mersenne primes. If n is proven an
integer then p + d is prime, therefore,
( + )
( + )
p + d = 2𝑝 𝑘 𝑚 – 1, is prime and 2𝑝 𝑘 𝑚 – 1,
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is a Mersenne prime outside our finite set of k Mersenne primes, therefore, our
assumption of a finite set of Mersenne primes is contradicted and it is proven that there
are an infinite number of Mersenne primes. It suffices to prove n is an integer.
8
((
𝑘 + 𝑚) − 2𝑝(𝑘) 𝑝(𝑘 + 𝑚) − 𝑝(𝑘)
(−1) 2𝑝(
�!
�2
2
1
1
1
(
)
𝑛 = (2𝑝 𝑘 – 1 − 1)! � 𝑝(𝑘 )
+
� + 𝑝 (𝑘 )
+
𝑘)
𝑘 + 𝑚)
(𝑘)
(
𝑝(
𝑝
𝑝
2
–1
2
– 1 +2
– 2
2
– 1 2𝑝(𝑘 )– 1 + 2𝑝 �𝑘 + 𝑚� – 2𝑝(𝑘)
Reducing,
+
( )
Multiplying by 2𝑝 𝑘 – 1,
𝑘 𝑚) − 𝑝(𝑘)
�2𝑝(
�!
2
1
1
1
(𝑘 )
𝑝
𝑛 = (2
– 2)! � 𝑝(𝑘 )
+
+
𝑘 + 𝑚 ) � + 𝑝 (𝑘 )
𝑘 + 𝑚)
(
𝑝
𝑝
2
–1
−1 + 2
2
– 1 −1 + 2 (
+
( )
𝑝(𝑘 ) – 1)�2𝑝(𝑘 𝑚) − 2𝑝(𝑘)�!
(2𝑝 𝑘 – 1)
(2
(
)
(
)
𝑝
𝑘
𝑝
𝑘
+
𝑛(2
– 1) = (2
– 2)! �1 +
�+1+
𝑘 𝑚)
𝑘 + 𝑚)
1 + 2𝑝(
−1 + 2𝑝(
( + )
Multiplying by (2𝑝 𝑘 𝑚 - 1),
𝑘
( )
𝑘
( + )
( + )
( + )
𝑛(2𝑝( )– 1)(2𝑝 𝑘 𝑚 – 1) = (2𝑝 𝑘 – 2)! �2𝑝 𝑘 𝑚 – 1) + (2𝑝 𝑘 𝑚 – 1)�! 2𝑝(
+𝑚 −
) 2𝑝(𝑘)�!� + 2𝑝(𝑘 + 𝑚) – 1 + (2𝑝(𝑘 ) – 1)
Reducing again,
𝑘
( )
𝑘
( + )
( + )
( + )
𝑛(2𝑝( )– 1)(2𝑝 𝑘 𝑚 – 1) = (2𝑝 𝑘 – 2)! �2𝑝 𝑘 𝑚 – 1) + (2𝑝 𝑘 𝑚 – 1)�! 2𝑝(
9
+𝑚 −
) 2𝑝(𝑘)�!� + 2𝑝(𝑘 + 𝑚) + (2𝑝(𝑘 ) – 2)
We already know p is prime, therefore, p = integer. Since p is an integer and by
definition k and m are integers, the right hand side of the above equation is an integer.
Since the right hand side of the above equation is an integer and p and k are integers
on the left hand side of the equation, then n must be an integer for the left side of
equation to be an integer, or n would need to be a rational fraction that is divisible by p.
x
x
This implies that n = where, p is prime and x is an interger. Then p = , since p is
p
n
prime, then p is only divisible by p and 1, therefore, n can only be equal to p or 1 in this
case, which are both integers, thus n must be an integer. It suffices to show that n is an
( )
( + )
integer since we have proven that p = 2𝑝 𝑘 – 1 and p + d = 2𝑝 𝑘 𝑚 – 1, where d =
( + )
( )
2𝑝 𝑘 𝑚 - 2𝑝 𝑘 , are both primes if and only if for integer n:
1 (−1)dd!
1
1
𝑛 = (𝑝 − 1)! � +
�+ +
𝑝
p + d
p p+d
( )
( + )
Since n = integer, we have proven that p = 2𝑝 𝑘 – 1 = 𝑀𝑝(𝑘) and p + d = 2𝑝 𝑘 𝑚 – 1
( + )
are both prime. Since we proved earlier that if 2𝑝 𝑘 𝑚 – 1 is prime then it also is not in
()
the finite set of 2𝑝 𝑖 – 1 Mersenne primes, therefore, since we have proven that
( + )
2𝑝 𝑘 𝑚 – 1 is prime, then we have proven that there is a Mersenne prime outside
the our assumed finite set of Mersenne primes. This is a contradiction from our
assumption that the set of Mersenne primes is finite, therefore, by contradiction the set
of Mersenne primes is infinite. Also this same proof can be repeated infinitely for each
finite set of Mersenne primes, in other words a new Mersenne prime can added to each
set of finite Mersenne primes. This thoroughly proves that an infinite number of
Mersenne primes exist.
10
References:
1) TYCM, Vol. 19, 1988, p. 191
2) 1001 Problems in Classical Number Theory, Jean-Marie De Koninck and Armel
Mercier, 2004
11