
#1
Jan2211, 09:53 AM

P: 66

THE PROOF THAT AN ODD PERFECT NUMBER DOES NOT EXIST
Let us firstly suppose that such an odd number (X) member of positive integers exist. We know that such a number couldnt be prime number, so lets factorize this number to prime factors. X = p1*a (1) where p1 is the lowest prime factor included in the equation and a all the other factor(s) arising from X. ( there could be another p1 or more in a, this case makes no problem for the proof.) Since we know that when we subtract a from (1) all the other combinations of factors have to sum so (p1 1 )*a = (1 + p1) + (1 + p1)*(k) (2) k... summed all of the combinations of factors in a, NOT a single 1 is restricted, but a itself not included, because it was already subtracted from X. Since when we have the prime number p1, (p1 – 1) and (p1 + 1) have just 1 common factor and that is the number 2. (Since when one of them is divisible by 3 the other is not and when 1 of them is divisible by 5, the other is not, and when one is divisible by seven, the other is not....) So when we rearrange (2) we get: (p1 – 1)*a = (p1 + 1)*(k + 1) (3) special case of the equation, when no combinations arising from a are restricted. since (p1 – 1) and (p1 + 1) they dont have common factors beside 2, it follows that (p1 + 1) has to include one or more factors of a. But since p1 is the lowest prime factor in X and since (p1 + 1) includes prime factors that are smaller than p1, we got a contradiction. We know, that if a = p2*a2 where p2 is the second smallest prime factor in X we will get maximum if we sume all of the combinations of factors in a (no factor is restricted) and we know that this sum k is a/(p2) < k we will show this maximum can never reach the condition (3) or (4) for odd numbers. (p1 – 1)*a < (p1 + 1)*(k + 1) (4) 4 holds true if we have a restricted combinations in equation (3), so the right part is greater if we include in (3) all of the combinations. Since a/(p2) < k if we include this into inequality we get: (p1 – 1)*a  (p1 + 1) < (p1 + 1)*(a/(p2)) p2*(p1 – 1)*a  (p1 + 1)*p2 < (p1 + 1)*(a) since p1 + 1 is > p1 we can write: p2*(p1 – 1)*a  (p1 + 1)*p2 < (p1 )*(a) p2/p1*((p1 – 1)*a – (p1 + 1)) < a (5) (p1 – 1)*a – (p1 + 1) = p1*a – a – p1 – 1 (6) If X is even and so p1 > 2 than (6) > a so we got a contradiction in (5). When p1 is 2, than p1*a – a – p1 – 1 < a so the solution of 5 may exist. So from this proof the perfect odd number does not exist and the even number may exist. The end of the proof. I could be totally wrong, but I found this Math problem yesterday. as with FLT it could be the right way how to prove it. Please respect I am just an amateur Mathematician 



#2
Jan2211, 10:19 AM

P: 66

I have forgotten to state that p2 could be greater OR equal to p1.




#3
Jan2311, 07:09 AM

P: 66

Could somebody improve the Math formulations and mistakes please?




#4
Jul2811, 09:32 AM

P: 66

The proof that an odd perfect number does not exist
REVISED:
If X is ODD and so p1 > 2 than (6) > a so we got a contradiction in (5). When X is EVEN, so p1 is 2, than p1*a – a – p1 – 1 < a so the solution of 5 may exist. So from this proof the perfect odd number does not exist and the even number may exist. 



#5
Jul3011, 10:01 AM

P: 688

Hi, Robert, one question,
what exactly did you do to arrive at eq.2? "a" was a collection of factors multiplied together, so where do the sums in the righthand side come from, and who is "k" exactly? 



#6
Aug211, 02:47 AM

P: 66

Dear Dodo, thanks for your interest. Its a rather clumsy proof, But I started it from the fact, that IF a such perfect number exists, if we factorize it to primes, ALL the possible combinations of multipled primes fe: p1*p2*p3 gives the combinations p1*p2, p2*p3, p1*p3, p1, p2,p3 and 1. Those are all divisors of the original number. If we sum them together ALL the possible combinations of those should sum to a original number OR greater if some combinations are restrictedread the same: as in number 28. We can easily see that if the first smallest prime is 5, those coombinations will NEVER sum to the near of original number. (But this is not included in the proof.) So the proof should be a lot shorter if we include this fact. Kind regards Robert




#7
Aug211, 03:04 AM

P: 66

In other words, from the proof we shall see that the necessary condition of existance of a perfect number is : Its smallest prime factor eqauls to 2. Cheers, Robert
ps:the only exceptions to this rule as I see it are 3*3*3 or 3 on n power, 3*3*5 and 3*3*7 (the sum of all the possible  double combinations included) sum up to a number greater than original one.. But when dealing with 3 on n power the sum of all DIFFERENT combinations never exceeds the original number. I believe with those conditions as I stated them, the proof should be easy to find. EDITED:It seems there are other combinations of n powers which exceed the original number when n is greater than the number to power; we can see that all the combinations sum up larger. It seems from the thinking, that this proof if only valid for the small numbers. This thinking leads us to the position if such an odd perfect number exists, the number of prime factors included must be equal or greater than the smallest prime, this are the boundary conditions for such a proof. anyway, Ill get back to this proof soon, its been a while. 


Register to reply 
Related Discussions  
Nonempty perfect set in R with no rational number  Calculus & Beyond Homework  8  
Finding Perfect Number with C++  Programming & Computer Science  5  
Perfect number  Linear & Abstract Algebra  6  
Smallest number to get a perfect square  Precalculus Mathematics Homework  7  
A new Perfect number formula  Linear & Abstract Algebra  9 