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May 23rd, 2018, 01:46 AM  #1 
Member Joined: Nov 2011 From: Vietnam Posts: 41 Thanks: 0  A proof of there is no odd perfect number A PROOF OF THERE IS NO ODD PERFECT NUMBER INTRODUCTION A perfect number is equal to the sum value of its positive divisors excluding itself. Perfect number is usually denoted P. Currently only even perfect numbers are to be known. It is not known if any odd perfect numbers exist. Here are several even perfect numbers and theirs explanations: 6 = 1+2+3 =2*3 28 = 1+2+4+7+2*7 = 4*7 496 = 1+2+4+8+16+31+2*31+4*31+8*31 = 16*31 8128 =1+2+4+8+16+3264+127+2*127+4*127+8*127+16*127+32*127= 64*127 By looking at these bold numbers, in general we rule right out the formula for even perfect number: $\displaystyle P = 2^{n1} (2^{n}1)$ (1) This is the general formula of a perfect number P formed by two primes $\displaystyle 2$ and $\displaystyle 2^{n}1$, when $\displaystyle 2^{n}1$ is a Mersenne prime. PROOF Looking at an only explanation form of the perfect number 8128: 8128 =1+2+4+8+16+32+64+127+2*127+4*127+8*127+16*127+32*127= 64*127 In general, a perfect number P (even or odd) to be made from two primes p and q, with q > p, then it has to be explanated as the perfect number 8128, and as follows: $\displaystyle P=1+p+p^{2}+p^{3}+...+p^{n1}+q+pq+p^{2}q +p^{3}q+...+p^{n2}q = p^{n1}q $ (2) $\displaystyle P = p ^{n1} q $ (2*) From the equation (2) we count q by p: $\displaystyle 1+p+p^{2}+p^{3}+...+p^{n1} = p^{n1}q(q+pq+p^{2}q+p^{3}q+... +p^{n2}q)$ $\displaystyle 1+p+p^{2}+p^{3}+...+p^{n1} = [p^{n1}(1+ p + p^{2} + p^{3} +...+ p^{n2})]q $ Add and minus with the same value $\displaystyle p^{n1} $ into the parentheses on the right side and reduce: $\displaystyle 1+p+p^{2}+p^{3}+...+p^{n1} = [p^{n1}+p^{n1}p^{n1}(1+p+p^{2}+p^{3}+...+p^{n2})]q$ $\displaystyle 1+p+p^{2}+p^{3}+...+p^{n1} = 2p^{n1} (1+p+p^{2}+p^{3}+...+ p^{n2}+ p^{n1})q $ $\displaystyle 1+p+p^{2}+p^{3}+...+p^{n1} = 2p^{n1}  (1+p+p^{2}+p^{3}+...+p^{n1})q $ From there: $\displaystyle q = \frac{1+p+p^{2}+p^{3}+...+p^{n1}} {2p^{n1}(1+p+p^{2}+p^{3}+...+p^{n1})}$ Multiply the numerator and the denominator with the same value p1 and simplify: $\displaystyle q = \frac{ (p1) (1+p+p^{2}+p^{3}+...+p^{n1})} {(p1)2p^{n1}(p1)(1+p+p^{2}+p^{3}+...+p^{n1})}$ $\displaystyle q = \frac{p^{n} 1} {2p^{n} 2p^{n1}(p^{n}1)}$ $\displaystyle q = \frac{p^{n} 1} {2p^{n} 2p^{n1}p^{n}+1}$ $\displaystyle q = \frac{p^{n} 1} {p^{n}2p^{n1}+1}$ From there we count q by p: $\displaystyle q = \frac{p^{n} 1} {p^{n}2p^{n1}+1}$ This is a formula noted the relation of the two primes p and q in the form of any perfect number P. Change $\displaystyle q = \frac{p^{n} 1} {p^{n}2 p^{n1}+1}$ to (2*), we have: $\displaystyle P = (p^{n1})*\frac{p^{n} 1} {p^{n}2p^{n1}+1}$ (3) This is the general formula of any perfect number P (even or odd) when p and q (that $\displaystyle q = \frac{p^{n} 1} {p^{n}2p^{n1}+1}$ ) are primes. The formula (2) or (2*) for perfect number to be made from two primes p and q, now is totally counted by the value of the only prime p in the formula (3) . $\displaystyle P=1+p+p^{2}+p^{3}+...+p^{n1}+q+pq+p^{2}q +p^{3}q+...+p^{n2}q = p^{n1}q $ (2) Or $\displaystyle P = p ^{n1} q $ (2*) → $\displaystyle P = (p^{n1})*\frac{p^{n} 1} {p^{n}2p^{n1}+1}$ (3) When p = 2, change to (3) we have: $\displaystyle P = 2^{n1}*\frac{2^{n} 1} {2^{n}2*2^{n1}+1}$ Because of $\displaystyle 2^{n}= 2*2^{n1}$, then the denominator $\displaystyle 2^{n}2*2^{n1}+1=2^{n}2^{n}+1=1 $, we have: $\displaystyle P = 2^{n1} (2^{n}1) $ From there we have again the formula for even perfect number $\displaystyle P = 2^{n1} (2^{n}1) $ , when $\displaystyle 2^{n}1$ is a Mersenne prime. When 3 ≤ p, is there or not a prime q with q > p that $\displaystyle q = \frac{p^{n} 1} {p^{n}2p^{n1}+1}$, along with the prime p in order to form an odd perfect number? To prove this, herein below are some little changes: Numerator: $\displaystyle p^{n} 1= p* p^{n1}1$ Denominator: $\displaystyle p^{n} 2p^{n1} +1 = p*p^{n1}2p^{n1} +1 = (p2)* p^{n1} +1 $ So $\displaystyle q = \frac{p^{n} 1} {p^{n}2 p^{n1}+1}$ to be changed by: $\displaystyle q=\frac{p*p^{n1} 1} {(p2)*p^{n1}+1}$. Let $\displaystyle p^{n1} = x $, we have: $\displaystyle q=\frac{p*p^{n} 1} {(p2)*p^{n1}+1}$ to be changed by: $\displaystyle q= \frac{px 1} {(p2)x+1} $ We see that: $\displaystyle n → ∞, p^{n1} → ∞, x → ∞ $. So: $\displaystyle q = lim x → ∞ \frac{p*p^{n} 1} {(p2)*p^{n1}+1}$ $\displaystyle q = lim x → ∞ \frac{px 1} {(p2)x+1}$ $\displaystyle q = lim x → ∞\frac{(px1)'}{((p2)x+1)'}$ $\displaystyle q = \frac{p}{p2}$ From there when $\displaystyle n → ∞ $, then $\displaystyle q = \frac{p}{p2}$ When $\displaystyle p=3, n → ∞, q → 3 $ … When $\displaystyle p=5, n → ∞, q → \frac{5}{3}$ … When $\displaystyle p=7, n → ∞, q → \frac{7}{5}$ … When $\displaystyle p=11, n → ∞, q → \frac{11}{9}$ … … When $\displaystyle p → ∞, n → ∞, q → 1 $ … q < p and q has no value of a prime larger than p. CONCLUSION As a result, there is not a prime q along with the prime p in order to form an odd perfect number P. SO THERE IS NO ODD PERFECT NUMBER Author: ĐOÀN ĐỨC NHUẬN by Vietnamese Nhuan Doan = Ducnhuandoan = Lawsfromabcmaths 
May 23rd, 2018, 06:22 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
Most of that I simply couldn't follow but you appear to conclude "As a result, there is not a prime q along with the prime p in order to form an odd perfect number P." But how does that imply "SO THERE IS NO ODD PERFECT NUMBER"? An odd number can be written as a sum of an even number and an odd number they don't have to be primes.. 
May 23rd, 2018, 05:43 PM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 621 Thanks: 394 Math Focus: Dynamical systems, analytic function theory, numerics 
Its nice to have a new flavor of crazy around here.


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