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Old 2010-02-03, 17:13   #166
blob100
 
Jan 2010

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I want to thank everyone who answered my questions and helped me.
Big thanks Tomer.
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Old 2010-02-03, 17:16   #167
R.D. Silverman
 
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Quote:
Originally Posted by CRGreathouse View Post
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"
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Old 2010-02-03, 18:46   #168
CRGreathouse
 
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Quote:
Originally Posted by R.D. Silverman View Post
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.
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Old 2010-02-04, 11:11   #169
blob100
 
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what is the tiger challenge?
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Old 2010-02-04, 15:16   #170
CRGreathouse
 
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Quote:
Originally Posted by CRGreathouse View Post
Quote:
Originally Posted by R.D. Silverman View Post
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.
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Old 2010-02-04, 21:09   #171
davieddy
 
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Quote:
Originally Posted by blob100 View Post
what is the tiger challenge?
Pull a bird
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Old 2010-02-04, 22:04   #172
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Quote:
Originally Posted by blob100 View Post
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)

Last fiddled with by cheesehead on 2010-02-04 at 22:06
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Old 2010-02-06, 11:47   #173
blob100
 
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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 2p-1.
A\equiv1\(mod\: p).

If 2p-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
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Old 2010-02-06, 12:20   #174
Mini-Geek
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Quote:
Originally Posted by blob100 View Post
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 2p-1
.
Shouldn't that just be "A is a factor of 2p-1", or "A is in the set..."?
Quote:
Originally Posted by blob100 View Post
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 View Post
If 2p-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 "2p-1 divides 2p-1". And that's certainly not restricted to a prime number, or a number of the form 2p-1.
Quote:
Originally Posted by blob100 View Post
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 View Post
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.

Last fiddled with by Mini-Geek on 2010-02-06 at 12:21
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Old 2010-02-06, 12:50   #175
blob100
 
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[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
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Old 2010-02-16, 15:06   #176
blob100
 
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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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