 # My Research on Perfect Numbers -Bhogapurapu Nageswara Rao

My Research on Perfect Numbers

Bhogapurapu Nageswara Rao

[13/7/20]

Natural Numbers and Classification

All natural numbers except 1 are either prime numbers or Multiples of Prime numbers.

We know that there are two types of numbers, those are even numbers and odd numbers.

I classified even numbers as

1. Pure Even Numbers
2. Impure Even Numbers

And again, odd numbers as

1. Pure Odd Numbers
2. Impure Odd Numbers

Pure Even Numbers:

If a number in the form of 2a, a ∈ N then it is called as pure even number. eg: 21 =2, 22 =4, 23 =8 …….
(It is like a female, who has only one type of chromozome)

Impure Even Numbers:

If 2a is multiplied with (polluted with) any odd prime or with odd primes then it is called as “Impure even number”. eg: 21(3)2 => 2(9)=18, 22(3)1 (5)2 => 12(25)=300
(It is like a male, who has two types of chromozomes)

Pure Odd Numbers:

If a number is in the form of Pa, where P is an odd prime number and a ∈ N. Then it is called “Pure odd number”.

Impure odd numbers:

If a number like Px a X Pzb X Pn c ……. where Px , Py , Pz   are various odd prime numbers(Odd Prime Numbers and a, b, c are ∈ N, are a=b=c or ab=c or a=bc or abc.

So Impure odd numbers are the product of “Different pure odd numbers”.

eg: 31X51 => 15, 32 X51 =>45, 71X31X112=2541.

Perfect number – My views

By the definition of Perfect number, we know that “A positive integer that is equal to the sum of its positive divisors excluding the number itself.

(OR)

If a double of a positive integer that is equal to the sum of its positive divisors then the positive integer called as “Perfect number”.

eg: divisors of 6 = {1,2,3,6} σ (6) = 12 => 2 X 6

Euclid proved that (2p-1) (2p-1), where P is a “Mersenne prime number” is a Perfect number.
Euler prove that all even perfect numbers of this form. So (2p-1) (2p-1) is called Euclid-Euler theorem.
6, 28, 496 and 8128 … are some perfects.

“It is not known that whether there are any odd perfect numbers”, by this statement we came to know that we don’t know whether there exist any odd perfect numbers in positive integers.

I proved that “there is no perfect number in the set of pure evens” or “in the set of pure odds”

Theorem

Any “pure even” or “pure odd” may not be perfect numbers”.

Theorem 1:

The sum of all positive divisors of a pure even number, never and can’t be expressed as ‘n’ times of the pure even number, where n∈N.

Proof:

Let us take a pure even number = Pa (where P=2 and a∈N)

We know that “the sum of all +ve divisors of Pa

σ(Pa) = Pa+1/P-1 [f(Pa) = +ve factors of Pa]

Can we express σ(Pa) = n. Pa ? n∈N… is it possible?

Let us check for it
Let us take that there is a possibility to express σ(Pa) = n. Pa

• Pa = Pa+1/P-1
• n = Pa+1 – 1/Pa(P-1)

LHS is ∈N, but RHS ∉N, why because Pa+1 – 1 is an odd (Pa+1  is an even), and Pa(P-1) is an even, we never get a number ∈Z+ when an odd number divisible by an even number.

• Pa+1 – 1/Pa(P-1) ∉ N

[ An odd number never purely divisible by an even number.]

Here we get contradiction => n ∈N, and n ∉N.

There is no “n” ∈N, to express as n.Pa = Pa+1 – 1/P-1

i.e σ(Pa) never be expressed “n” (n∈N) times of Pa.

We know that if we express σ(Pa)=2Pa then “Pa” is called as perfect number, here we proved that n∉N, so “any pure even number” may not be perfect number.

Theorem 2:

“There is no any pure odd number, that might be Perfect Number”.

The sum of all +ve divisors of a pure odd number and can’t be expressed as “n” times of the pure odd number, where n∈N.

Proof:

Let us take a pure odd number = Pa (where P is an odd prime number and a ∈N).

W.K.T “The sum of all _ve divisors of Pa

=> σ(Pa) = Pa+1 -1/P-1 [here σ(Pa) = sum of all +ve divisors of Pa]

Now the theorem is … “is it possible to find n∈N?”

=>σ(Pa) = n. Pa, n∈N,

If Pa+1 -1/P-1 is a pure divisor then we get n ∈N.

So, we have to prove that Pa+1 -1/P-1 is not a pure division.

If Pa+1 -1/P-1 is a pure division it must satisfy the “3 Rules of Pure division”.

Now let us take =>σ(Pa) = n. Pa, n∈N,

LHS    RHS

• n = σ(Pa)/ Pa if RHS is a pure division then we can find the existance of “n” in “N”

if RHS is not a pure division then we cannot find the existance of ‘n’ in N.

Now let us check whether RHS is pure divison or not,

RHS => σ(Pa)/ Pa

W.K.T. σ(Pa) = Pa+1 -1/P-1

σ(Pa)/ Pa = Pa+1 -1/ Pa (P-1) [Numerator Pa+1 -1 = s; Denominator Pa (P-1) =t]

Let us check that this s/t follow Rule 1 or not.

Rule 1 is s>t

• Pa+1 -1> Pa (P-1)
• Pa+1 -1-Pa+1 + Pa>0

To show s>t, we better to show that (i) st (ii) s not < t.

To show (i)st

Let us take s=t

Then => Pa+1 -1= Pa (P-1)

• Pa+1 -1= Pa+1 – Pa
• -1 = – Pa

1 = Pa [W.K.T P is an odd prime and a∈N. ∵ 1  Pa]

But we get a contradiction that 1=Pa

So, s=t is false ∵ st is True

(ii) We have to show s not <t

Let us take that s<t

• Pa+1 -1< Pa (P-1)
• Pa+1 -1< Pa+1 – Pa
• -1<- Pa
• 1> Pa Clearly, we can say that this is not true, because W.K.T. Pa>1

∴ We get a contradiction ∴ s not <t

From (i) and (ii) st and s not < t. ∴ s>t

So, the rule 1 Tested and satisfied.

Note we can show s>t? sure Let us take

• Pa+1 -1> Pa (P-1) => Pa+1 -1> Pa+1-Pa=>-1>-Pa=>1<Pa it is true.

∴ s>t.

To check Rule 2 => We have to show s≥2t

We have to check whether s&t satisfy Rule 2

=> Let us take s≥2t. If we did not get contradiction, we can say that s & t satisfy the rule 2.
Now let us take s≥2t

• Pa+1-1≥2[Pa(P-1)]

≥2 Pa+1 – 2 Pa

• 2 Pa -1≥2 Pa+1 – Pa+1
• 2 Pa-1 ≥ Pa+1
• 2 Pa – 1 ≥ Pa.P

Here, we get a contradiction 2 Pa -1 is never ≥ Pa.P

∵ we know that P is an odd Prime number => P≥3

Here, W.K.T 2<P where P is any odd prime

2Pa<P.Pa

∴ 2Pa-1 << P.Pa

∴We get s not ≥ 2t

∴ s<2t So here s,t do not satisfy the rule 2. So, s&t fails here to satisfy pure division Rules.

∴ We cannot get n=σ(Pa)/ Pa, where n∈N

∴ We can’t express that the sum of +ve divisors of a pure add number ….

Never and ever express as n times of the pure odd number.

So, we can say darely that there is no “perfect pure odd number”

(∵ if n=2 then it is called perfect number).

Trifect numbers “n”fect numbers

-my Thesis

[Bhogapurapu Nageswara Rao]

What is Trifect number?

We know that a perfect number is double of itself equals to the “sum of its positive divisors”. Let us call this as “Duofect Number” or “Twofect number”

As like as ablove definition if a number Thrice (three times) of itself equals to the “sum of its +ve divisors”, let us call the number as “Trifect number”.

I find “120” is a Trifect number and it is the smallest, first Trifect number,

Are there any other Trifect numbers?

Why because 120 factors = {1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120}

σ(120) = 360 => 3(120)

I don’t know whether there exist any other Trifect number or not.

Note: The word “Trifect number” created by me.

Conclusion

“Are there any odd Perfect numbers?” This is an unsolved question in Mathematics till today.

Though I didn’t solve the question completely, I did solve it partially. With this research, I have also extended the idea of Perfect number as Trifect number, fourfect number… and as “n”fect number (n∈N).

My major findings in this research are:

• There is no perfect number in pure even numbers or in pure odd numbers.
• And there is no “n”fect number in pure even numbers or in pure odd numbers.

For this paper I invented the following Terms:

1. Pure even numbers
2. Impure even numbers
3. Pure odd numbers
4. Impure odd numbers
5. Pure divisions
6. Impure divisions
7. Trifect numbers
8. “n”fect numbers CLOSE