In a message dated 4/29/00 6:31:10 AM Pacific Daylight Time,
[EMAIL PROTECTED] writes:
<< Good. Now that we are on the topic of perfect numbers, I have proven that
any odd perfect number must have at least 9 factors. Another boy at my
school is trying to tell me that all perfect numbers are in the from
(a^(p-1))(a^p-1), thus showing that there are no odd perfect numbers. Can
someone please explain this to me? I am quite sure that he is wrong.
>>
The kid at school probably was talking about even perfect numbers. (a = 2)
Anyway, the proof I had goes like this, in case you are wondering, because,
like I said, I was pretty busy.
Anyway: Here are a couple of things you need to know.
(2(n) + 1) is the formula for odd numbers. (2^(n) - 1) is always odd, (2^n -
2) is an odd times 2. Another is that even + even = even, and even + odd =
odd. There is also this Pascal's triangle.
64
32 32
16 32 16
8 24 24 8
4 16 24 16 4
2 10 20 20 10 2
1 6 15 20 15 6 1
You'll notice that this is the same as the regular triangle, except you
multiply powers of two. Now for the proof:
_________________________________________________
By definition of an odd, perfect number, with only distinct factors, the
following must be true, if the odd, perfect number has only two factors:
(2)(2a + 1)(2b + 1) = (2a + 1)(2b + 1) + (2a + 1) + (2b + 1) + 1
(2a + 1)(2b + 1) = 2(a + b) + 2 + 1
4ab + 2(a + b) + 1 = 2(a + b) + 2 + 1
4ab = 2
2ab = 1
This is the same as saying (even = odd), which is an impossibility.
If you go to three factors, four factors, or even five factors, you'll notice
a pattern, having to do with the triangle above. You keep on dividing the
starting number in half for that triangle. Watch with 6 factors how it
relates to the triangle above.
X1 = sum of combinations formed from 6 (abcdef)
X2 = sum of combinations formed from 5 (abcde+abcdf+abcef+etc)
X3 = sum of combinations formed from 4 (abcd+abce+etc)
...
X6 = sum of combinations formed from 1 (a+b+c+d+e+f)
64 * X1 + 32 * X2 + 16 * X3 + 8 * X4 + 4 * X5 + 2 * X6 + 1 =
(32(X2) + 32(X3) + 24(X4) + 16(X5) + 10(X6) + 6) + (16(X3) + 24(X4) +
24(X5) + 20(X6) + 15) + (8(X4) + 16(X5) + 20(X6) + 20)
+ (4(X5) + 10(X6) + 15) +(2(X6) + 6) + 1.
Notice the similarities between this and the Pascal's triangle? (write the
triangle if necessary) Now, every row of that triangle equals 64, because of
the reason in the e-mail sent by the other guy. Anyway, after subtracting,
from the above equation gives the middle of the triangle:
32
24 24
16 24 16
10 20 20 10
6 15 20 15 6
And the rows add up to: (64 - 64), (64 - 32) , (64 - 16), (64 - 8), (64 - 4)
, and (64 - 2), and this was proven by the last guy's email, since the first
triangle is just the regular triangle times descending powers of 2:
1 * 64 = 64
1 1 * 32 = 32 32
1 2 1 * 16 = 16 32 16
1 3 3 1 * 8 = 8 24 24 8
etc..
So, this goes to:
64(X1) = 32(2-1)(X3) + 16(4-1)(X4) + (8)(8-1)(X5) + 4(16-1)(X6) + 2(32-1) * 1
This can be divided by two, giving:
32(X1) = 16(2-1)(X3) + 8(4-1)(X4) + 4(8-1)(X5) + 2(16-1)(X6) + (32-1)(1)
which is saying:
even = even + even + even + even + odd
even = even + odd
even = odd
Which is a good enough contradiction.
I explained a whole lot rather than just doing algebra, because we can take
the same route for the general equation rather than multiplying out, however,
I think it may have several problems, because it might not apply to the
general equation. Any way, I hope this helps. Maybe someone can do the
rest, and so the whole world can confidently say that if an odd, perfect
number exists, then it must be 0 mod ((2n+1)^2). Then we can go on to
squares times odd numbers, cubes time odd numbers, etc. I just hope the
whole thing is provable, after this.
Hoping and praying that 2^(9025267) - 1 is prime!!!!!!!
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