06-11-2008, 06:21 AM
B"H
I will post a few more threads before I have to tidy things up and move on. Here is an excerpt of an attempted proof that even perfect numbers are 1(mod9). I sent this to Ask Dr. Math a while ago. I rather suspect that it has been thought of before, given that the good folks at Drexel U. seem to have shelved it. It is a rather elementary proof.
However, I will still ask for a critique from the list. After I have posted my last post, I will remain a while for PM's and people who want to correspond.
Here is the excerpt. One more thing, I *REALLY* apologize for my lack of knowledge as to proper number theory language. I a not formally educated in it. So, I will take all corrections on dictum with due humility. OK, here it is...
"Is this a proof that an even perfect above 6 number is 1(mod9)? An
even perfect number formula:
[2^p-1][(2^p)-1], in which the latter term is a Mersenne.
OK, well, if we let 2^p-1 be defined as 1modp (Fermat's little
theorem). Let that be defined as (x*p) +1. And, any even p-1 would
be an exponent of 2^p-1, which would still be 1modp. Now, 2^p would
be 2 times the value of 2^p-1, which would be (2x*p)+2. Now, (2^p)-1
would be 2x*p+1, which would also be 1modp. However, notice what
happens when we multiply values
(x*p + 1)(2x*p +1). We get (2x^2)p^2 + 3x*p +1.
Now, if we begin with the prime number 3, letting p=3, we can see that
this will produce 1mod9. That is so because 3x*p is 3x*3, which is
9*x, a multiple of nine. p^2 is 3^2, nine. That leaves us with some
multiple of nine plus a one.
OK, we can see that for any 2^p-1, as an even number it will be an
exponent of 2^2, which is 2^3-1. That means that 1mod3 for p=3
remains 1mod3 for any 2^p-1. And, we can see, therefore, that any p,
if fed in to the algorithm above, will produce 1mod9, provided that it
is an odd value for p and not p=2 (which gets us 6).
Therefore, if I am correct, we have an easily understandable way of
proving that even perfect numbers are 1mod9. Interestingly, it would
seem to suggest that other odd values h for [2^h-1][(2^h)-1] would be
1mod9 as well."
I will post a few more threads before I have to tidy things up and move on. Here is an excerpt of an attempted proof that even perfect numbers are 1(mod9). I sent this to Ask Dr. Math a while ago. I rather suspect that it has been thought of before, given that the good folks at Drexel U. seem to have shelved it. It is a rather elementary proof.
However, I will still ask for a critique from the list. After I have posted my last post, I will remain a while for PM's and people who want to correspond.
Here is the excerpt. One more thing, I *REALLY* apologize for my lack of knowledge as to proper number theory language. I a not formally educated in it. So, I will take all corrections on dictum with due humility. OK, here it is...
"Is this a proof that an even perfect above 6 number is 1(mod9)? An
even perfect number formula:
[2^p-1][(2^p)-1], in which the latter term is a Mersenne.
OK, well, if we let 2^p-1 be defined as 1modp (Fermat's little
theorem). Let that be defined as (x*p) +1. And, any even p-1 would
be an exponent of 2^p-1, which would still be 1modp. Now, 2^p would
be 2 times the value of 2^p-1, which would be (2x*p)+2. Now, (2^p)-1
would be 2x*p+1, which would also be 1modp. However, notice what
happens when we multiply values
(x*p + 1)(2x*p +1). We get (2x^2)p^2 + 3x*p +1.
Now, if we begin with the prime number 3, letting p=3, we can see that
this will produce 1mod9. That is so because 3x*p is 3x*3, which is
9*x, a multiple of nine. p^2 is 3^2, nine. That leaves us with some
multiple of nine plus a one.
OK, we can see that for any 2^p-1, as an even number it will be an
exponent of 2^2, which is 2^3-1. That means that 1mod3 for p=3
remains 1mod3 for any 2^p-1. And, we can see, therefore, that any p,
if fed in to the algorithm above, will produce 1mod9, provided that it
is an odd value for p and not p=2 (which gets us 6).
Therefore, if I am correct, we have an easily understandable way of
proving that even perfect numbers are 1mod9. Interestingly, it would
seem to suggest that other odd values h for [2^h-1][(2^h)-1] would be
1mod9 as well."