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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

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

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