aa r X i v : . [ m a t h . G M ] J a n In Memory of Maryam Mirzakhani (1977-2017)
THERE ARE NO ODD PERFECT NUMBERS
HOOSHANG SAEID–NIA
Abstract.
While even perfect numbers are completely characterized, the existence or otherwise ofodd perfect numbers is an open problem. We address that problem and prove that if a natural numberis odd, then it’s not perfect. introduction The following is one of the ancient open problems in number theory:
Definition. (Perfect Number) A natural number n is said to be perfect if the sum of all its (positive)divisors, including n itself, is equal to n . (1.1) X d | n d = 2 n. Example:
6, 28, 496, and 8128 are the first few perfect numbers. ✁ All even perfect numbers are completely determined [R1] by the following theorem:
Theorem.
Even Perfect Numbers A) Euclid (300 B.C.) if p − is prime, then n = 2 p − (2 p − is a perfect number. (Elements, BookIX, Proposition 36, as cited in [R1]). B) Euler (1707 – 1783) If n is even, then the converse of part (A) is also true. [R1] The ancient open problem is whether or not any odd perfect numbers exist? Section 2 of this inquiryis a concise review of the preliminaries. In section 3 we present a counter–proof for the existence of oddperfect numbers.A good account of previous work on this topic can be found, for example in [R2] and [R3].2.
Preliminaries
It’s understood that 1 is not a perfect number. The sum of divisors of 1 is 1, which is not two times1. Therefore, n > , and it can be uniquely factorized as n = m Y i =1 p a i i ( p i odd prime) ( a i positive).The sum of divisors of n is usually denoted by σ ( n ) . For a prime power, Date : January 19, 2021.2010
Mathematics Subject Classification.
Primary 11N25, Secondary 11Y50.
Key words and phrases. perfect numbers, odd perfect, Mersenne primes. OOSHANG SAEID–NIA σ ( p a ) = a X j =0 p j = p a +1 − p − .σ is a multiplicative function [R1]. So for every n in general, it can be written as:(2.1) σ ( n ) = m Y i =1 σ ( p a ) , Therefore, using (1.1), our main equation is(2.2) m Y i =1 σ ( p a ) = 2 m Y i =1 p a i i ( p i = 2 , a i > , m ≥ ).To solve the problem we have to exhibit a literal odd number with this property, or – as we do in thenext section – prove that it’s impossible.3. Odd perfect numbers don’t exist
Proposition.
If a natural number is odd, then it’s not a perfect number.Proof.
We argue by contradiction. Let n > be an odd number and assume, on the contrary to thestatement above, that n is perfect. Hence, 2.2 holds. For m = 1 , the solution is not difficult ([R1]).For m ≥ , we begin with the most basic observation, that only one σ ( p a i i ) must be even, and of theform S , with odd S . We may assume that i = 1 , possibly after relabeling the primes.(3.1) σ ( p a ) = 2 Sσ ( p a ) is a divisor of n , and it’s coprime to p a . So,(3.2) S = m Y i =2 p b i i , in which ≤ b i ≤ a i , with at least one b i > . We define(3.3) G = m Y i =2 p c i i , so that b i + c i = a i . Now we have(3.4) m Y i =2 p a i i = SG, andhere Are No Odd Perfect Numbers(3.5) m Y i =2 σ ( p a i i ) = p a G. Using these equations, we derive(3.6) p m Y i =2 σ ( p a i i ) = 2( p − m Y i =2 p a i i + G. To see this, simply start with p a +11 = ( p − σ ( p a ) + 1 , and multiply both sides by G . Slightly morecomplicated, but similarly we get(3.7) p p m Y i =3 σ ( p a i i ) = 2( p − p − m Y i =3 p a i i + 2( p a +11 + p a +12 − Q mi =3 σ ( p a i i ) σ ( p a ) σ ( p a ) . We define(3.8) G ′ = 2( p a +11 + p a +12 − Q mi =3 σ ( p a i i ) σ ( p a ) σ ( p a ) , so we can work with the more compact form of 3.7.(3.9) p p m Y i =3 σ ( p a i i ) = 2( p − p − m Y i =3 p a i i + G ′ . Now, multiply both sides of this by p a σ ( p a ) .(3.10) p a +11 p a +12 G = ( p − p − σ ( p a ) σ ( p a ) G + p a σ ( p a ) G ′ and get(3.11) G ′ = ( p a +11 + p a +12 − Gp a σ ( p a ) . The two equations 3.8 and 3.11 must agree. ( p a +11 + p a +12 − Gp a σ ( p a ) = 2( p a +11 + p a +12 − Q mi =3 σ ( p a i i ) σ ( p a ) σ ( p a ) Gp a σ ( p a ) = 2 Q mi =3 σ ( p a i i ) σ ( p a ) σ ( p a ) σ ( p a ) σ ( p a ) G = 2 p a m Y i =2 σ ( p a i i ) σ ( p a ) σ ( p a ) = 2 p a p a OOSHANG SAEID–NIASo, p a p a is a perfect number with two distinct primes. Therefore, contrary to our assumption, itmust be even. Alternately, if [ σ ( p a ) σ ( p a )] /p a p a is 2, then the product of it by σ ( p a i i ) /p a i i > term(s)is greater than 2. From this contradictions we conclude that an odd number n cannot be perfect. Thiscompletes the proof. (cid:3) Conclusion
We proved that perfect numbers are always even, and therefore, always built from Mersenne primes.Whether or not the set of Mersenne primes is infinite, is another interesting open problem [R3].
References [R1] John Voight,
Perfect Numbers: An Elementary Introduction ,http://math.dartmouth.edu/ ∼ jvoight/notes/perfelem.pdf[R2] Oliver Knill, The oldest open problem in mathematics ∼ knill/seminars/perfect/handout.pdf[R3] Richard K. Guy, Unsolved Problems in Number Theory , Springer, 2004.
HOOSHANG SAEID–NIA
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