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## Main Question or Discussion Point

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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