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

[QUOTE=CRGreathouse;204356]It's not a very modern treatment, but if you can find a copy of Underwood Dudley's [i]Number Theory[/i] it would certainly be easy enough.[/QUOTE]

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

[QUOTE=R.D. Silverman;204444]Everyone should read Dudley's "What to do when the trisector comes"
and his book "Mathematical Cranks"[/QUOTE]

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

what is the tiger challenge?

[QUOTE=CRGreathouse;204463][QUOTE=R.D. Silverman;204444]Everyone should read Dudley's "What to do when the trisector comes"
and his book "Mathematical Cranks"[/QUOTE]
I'm ashamed to admit I've read neither. I've checked out the latter to remedy that unfortunate gap in my reading. :smile:[/QUOTE]

OK, finished [i]Mathematical Cranks[/i]. Good book, very readable.

[quote=blob100;204544]what is the tiger challenge?[/quote]
Pull a bird

[quote=blob100;204544]what is the tiger challenge?[/quote]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. ([url]http://www.springsgreetingcards.com/catalogs/store.asp?pid=250913[/url])

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

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

[quote=blob100;204714]Here is another conjecture I tought about:
If [tex]2^{p}\equiv1\(mod\: A)[/tex].
Where p is an odd prime number,and [B]A is the set of the odd prime factors
of 2[sup]p[/sup]-1[/B].[/quote]
Shouldn't that just be "A is a factor of 2[sup]p[/sup]-1", or "A is [B]in[/B] the set..."?
[quote=blob100;204714][tex]A\equiv1\(mod\: p)[/tex].[/quote]
This is a result of the known fact that any factor of 2^p-1 is of the form 2kp+1.
[quote=blob100;204714]If 2[sup]p[/sup]-1 is a prime number, [tex]2^{p}\equiv1\(mod\:2^{p}-1)[/tex] exists.[/quote]
What? If I get what you're saying, that's equivalent to the trivial equation [tex]2^{p}-1\equiv0\(mod\:2^{p}-1)[/tex], or "2[sup]p[/sup]-1 divides 2[sup]p[/sup]-1". And that's certainly not restricted to a prime number, or a number of the form 2[sup]p[/sup]-1.
[quote=blob100;204714][tex]2^{p}-2\equiv0\(mod\:p)[/tex].
...
[tex]2^{p}\equiv1\(mod\:p)[/tex].[/quote]
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: [spoiler]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, [/spoiler][tex]a^{p-1} \equiv 1 \(mod\:p)[/tex]
[quote=blob100;204714]From here, the conjecture "says" that [tex]2^{p}-2\equiv0\(mod\:p)[/tex].
[tex]2(2^{p}-1)\equiv0\(mod\:p)[/tex].
[tex]2^{p}-1\equiv0\(mod\:p)[/tex].
[tex]2^{p}\equiv1\(mod\:p)[/tex].
Which is fermat's little theorem.[/quote]
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.

[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.

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, [tex]\sum_{n=1}^{\infty}n^{2}=1+3+5....2n-1[/tex]
[tex]\sum_{n=1}^{\infty}n^{2}=1+3+5....2n-1+2n[/tex]