how to know if my ideas didnt tought before? - Page 16

blob100 · 2010-02-03, 17:13

I want to thank everyone who answered my questions and helped me.
Big thanks Tomer.

R.D. Silverman · 2010-02-03, 17:16

Quote:

Originally Posted by CRGreathouse

It's not a very modern treatment, but if you can find a copy of Underwood Dudley's Number Theory it would certainly be easy enough.

Everyone should read Dudley's "What to do when the trisector comes"
and his book "Mathematical Cranks"

CRGreathouse · 2010-02-03, 18:46

Quote:

Originally Posted by R.D. Silverman

Everyone should read Dudley's "What to do when the trisector comes"
and his book "Mathematical Cranks"

I'm ashamed to admit I've read neither. I've checked out the latter to remedy that unfortunate gap in my reading.

blob100 · 2010-02-04, 11:11

what is the tiger challenge?

CRGreathouse · 2010-02-04, 15:16

Quote:

Originally Posted by CRGreathouse

Quote:

Originally Posted by R.D. Silverman

Everyone should read Dudley's "What to do when the trisector comes"
and his book "Mathematical Cranks"

I'm ashamed to admit I've read neither. I've checked out the latter to remedy that unfortunate gap in my reading.

OK, finished Mathematical Cranks. Good book, very readable.

davieddy · 2010-02-04, 21:09

Quote:

Originally Posted by blob100

what is the tiger challenge?

Pull a bird

cheesehead · 2010-02-04, 22:04

Quote:

Originally Posted by blob100

what is the tiger challenge?

If you mean "Year of the Tiger Challenge", it's

[Year of the Tiger] Challenge

In the traditional Chinese zodiacal calendar, a Year of the Tiger starts on Feb. 14, 2010. (http://www.springsgreetingcards.com/...asp?pid=250913)

blob100 · 2010-02-06, 11:47

Here is another conjecture I tought about:
If $2^{p}\equiv1\(mod\: A)$ .
Where p is an odd prime number,and A is the set of the odd prime factors
of 2^p-1.
$A\equiv1\(mod\: p)$ .

If 2^p-1 is a prime number, $2^{p}\equiv1\(mod\:2^{p}-1)$ exists.
From here, the conjecture "says" that $2^{p}-2\equiv0\(mod\:p)$ .
$2(2^{p}-1)\equiv0\(mod\:p)$ .
$2^{p}-1\equiv0\(mod\:p)$ .
$2^{p}\equiv1\(mod\:p)$ .
Which is fermat's little theorem.

Mini-Geek · 2010-02-06, 12:20

Quote:

Originally Posted by blob100

Here is another conjecture I tought about:
If $2^{p}\equiv1\(mod\: A)$ .
Where p is an odd prime number,and A is the set of the odd prime factors
of 2^p-1.

Shouldn't that just be "A is a factor of 2^p-1", or "A is in the set..."?

Quote:

Originally Posted by blob100

$A\equiv1\(mod\: p)$ .

This is a result of the known fact that any factor of 2^p-1 is of the form 2kp+1.

Quote:

Originally Posted by blob100

If 2^p-1 is a prime number, $2^{p}\equiv1\(mod\:2^{p}-1)$ exists.

What? If I get what you're saying, that's equivalent to the trivial equation $2^{p}-1\equiv0\(mod\:2^{p}-1)$ , or "2^p-1 divides 2^p-1". And that's certainly not restricted to a prime number, or a number of the form 2^p-1.

Quote:

Originally Posted by blob100

$2^{p}-2\equiv0\(mod\:p)$ .
...
$2^{p}\equiv1\(mod\:p)$ .

Take a look at these two "equivalent" statements and tell me if you see anything off.

You've gotta see it. What about where you went wrong?

Well? If you can't figure it out, highlight this text to see it: You're saying 2^p-2 is equivalent to 2^p-1, which is of course not true. The problem is caused by turning 2^p-2 into 2(2^p-1) instead of 2(2^(p-1)-1). If you correctly transform that, you get an alternate form of Fermat's little theorem, $a^{p-1} \equiv 1 \(mod\:p)$

Quote:

Originally Posted by blob100

From here, the conjecture "says" that $2^{p}-2\equiv0\(mod\:p)$ .
$2(2^{p}-1)\equiv0\(mod\:p)$ .
$2^{p}-1\equiv0\(mod\:p)$ .
$2^{p}\equiv1\(mod\:p)$ .
Which is fermat's little theorem.

So you're saying that your conjecture is equivalent to Fermat's little theorem, being a subset of it that adds nothing new? Umm...I think it should be pretty clear that the fact there is already known.

blob100 · 2010-02-06, 12:50

[quote=Mini-Geek;204717]
This is a result of the known fact that any factor of 2^p-1 is of the form 2kp+1.

I didn't know about this fact, so I found it by myself (somehow).
Thanks for the help.
BTW: I didn't see the part about fermat's little theorem was not correct, i meant what you wrote.

blob100 · 2010-02-16, 15:06

I have a new conjecture (it isn't about mersenne numbers),
I conjecture that: between n^2 to n^2+2n there is always a prime number.
When n is an odd number.
We can show n^2+2n as n(n+2).
So, $\sum_{n=1}^{\infty}n^{2}=1+3+5....2n-1$
$\sum_{n=1}^{\infty}n^{2}=1+3+5....2n-1+2n$

2010-02-06, 11:47	#173
blob100 Jan 2010 379₁₀ Posts	Here is another conjecture I tought about: If $2^{p}\equiv1\(mod\: A)$ . Where p is an odd prime number,and A is the set of the odd prime factors of 2^p-1. $A\equiv1\(mod\: p)$ . If 2^p-1 is a prime number, $2^{p}\equiv1\(mod\:2^{p}-1)$ exists. From here, the conjecture "says" that $2^{p}-2\equiv0\(mod\:p)$ . $2(2^{p}-1)\equiv0\(mod\:p)$ . $2^{p}-1\equiv0\(mod\:p)$ . $2^{p}\equiv1\(mod\:p)$ . Which is fermat's little theorem. Last fiddled with by blob100 on 2010-02-06 at 11:56

2010-02-06, 12:50	#175
blob100 Jan 2010 379 Posts	[quote=Mini-Geek;204717] This is a result of the known fact that any factor of 2^p-1 is of the form 2kp+1. I didn't know about this fact, so I found it by myself (somehow). Thanks for the help. BTW: I didn't see the part about fermat's little theorem was not correct, i meant what you wrote. Last fiddled with by blob100 on 2010-02-06 at 12:52

2010-02-16, 15:06	#176
blob100 Jan 2010 379 Posts	I have a new conjecture (it isn't about mersenne numbers), I conjecture that: between n^2 to n^2+2n there is always a prime number. When n is an odd number. We can show n^2+2n as n(n+2). So, $\sum_{n=1}^{\infty}n^{2}=1+3+5....2n-1$ $\sum_{n=1}^{\infty}n^{2}=1+3+5....2n-1+2n$ Last fiddled with by blob100 on 2010-02-16 at 15:10

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2010-02-03, 17:13	#166
blob100 Jan 2010 379 Posts	I want to thank everyone who answered my questions and helped me. Big thanks Tomer.

2010-02-04, 11:11	#169
blob100 Jan 2010 17B₁₆ Posts	what is the tiger challenge?