[QUOTE=blob100;213325]About the sum, I tought it is a known sign, as "e" for exponent or even "n" for natural, where it is about a set.
[/QUOTE]

'e' will always be assumed to be Euler's constant unless stated otherwise.
'n' is not generically used to mean 'natural number'. You have to
define it.

Learn to define your variables.

[QUOTE]
Is my proof correct?
To prove it by induction too right?[/QUOTE]

You have not given a proof.

[LEFT]I'm the one who is needed to be dissapointed because of me.
I solved just the first problem, and it took me like a month.
I can't solve the problem with the factor of fermat numbers.
And I ask you if you will be able to diffine [U]everything[/U] in the chinese reminder problem.

For the solvement:
1) for N=1:
F(1)=F(0)+1=2
We see that (N^2+N+2)/2=2.
2) we agree F(N-1)+N=(N^2+N+2)/2.
F(N+1)=F(N)+N+1.
We see that:
((N+1)^2+(N+1)+2)/2=(N^2+2N+1+N+1+2)/2=(N^2+3N+4)/2=
(N^2+N+2)/2+(2N+2)/2=F(N)+N+1=F(N+1).[/LEFT]

[QUOTE=blob100;213449][LEFT]I'm the one who is needed to be dissapointed because of me.
I solved just the first problem, and it took me like a month.[/QUOTE]

Silverman's problems are hard; don't feel bad if you have trouble with them. They're certainly at a higher level than the questions you'd get at school.

[quote=CRGreathouse;213476]Silverman's problems are hard; don't feel bad if you have trouble with them. They're certainly at a higher level than the questions you'd get at school.[/quote]
Of course these are harder then school's problems, eventually, school's problems are triviall.

[QUOTE=blob100;213491]Of course these are harder then school's problems, eventually, school's problems are triviall.[/QUOTE]

Shall we tackle problem 2?

Start with:

Let p be prime. any prime factor q of 2^p+1 is of the form q = 2kp+1.

Do you know how to show this? If 2^p+1 = 0 mod q, then 2^p = -1 mod q,
whence 2^(2p) = 1 mod q. Do you know any theorems about the order
of an element modulo a prime? Shanks' book will have discussed this.

Now, for 2^2^n + 1, it is easily seen that any factor must be of the form

2k(2^n) +1) by result above. We want to show that the factor must be of
the form 2k(2^(n+1) ) + 1. We know that this prime is 1 mod 8.
Now ask yourself: what relation is known between the prime 2, and primes
that are 1 mod 8? Now consider a certain result due to Euler.

[quote=R.D. Silverman;213498]Shall we tackle problem 2?

Start with:

Let p be prime. any prime factor q of 2^p+1 is of the form q = 2kp+1.

Do you know how to show this? If 2^p+1 = 0 mod q, then 2^p = -1 mod q,
whence 2^(2p) = 1 mod q. Do you know any theorems about the order
of an element modulo a prime? Shanks' book will have discussed this.

Now, for 2^2^n + 1, it is easily seen that any factor must be of the form

2k(2^n) +1) by result above. We want to show that the factor must be of
the form 2k(2^(n+1) ) + 1. We know that this prime is 1 mod 8.
Now ask yourself: what relation is known between the prime 2, and primes
that are 1 mod 8? Now consider a certain result due to Euler.[/quote]
About the first question, I don't know any theorem about order of a set.
If 2^(p)+1 is devided by q, 2^(2p)-1 is devided by q becuase:
(2^(p)-1)(2^(p)+1)=2^(2p)-1.

[QUOTE=blob100;213502]About the first question, I don't know any theorem about order of a set.
If 2^(p)+1 is devided by q, 2^(2p)-1 is devided by q becuase:
(2^(p)-1)(2^(p)+1)=2^(2p)-1.[/QUOTE]

I did not say anything about the order of a set. I did discuss order of
an element.

I assume that you know Fermat's little theorem.

What does it tell us about the order of an element?

[quote=R.D. Silverman;213527]I did not say anything about the order of a set. I did discuss order of
an element.

I assume that you know Fermat's little theorem.

What does it tell us about the order of an element?[/quote]
Ha, I tought you are asking about a set becuase of element is a part of a set, sorry.
Fermat little theorem assumes that:
2^(p-1)=1(mod p).
Or more generalized:
2^(phi(m))=1(mod m).
By order you mean?:
If e is the smallest exponent such that a^e=1(mod m) and (a,m)=1
Then so, a is of [U]order [/U]e modulo m?

[quote=blob100;213537]Ha, I tought you are asking about a set becuase of element is a part of a set, sorry.
Fermat little theorem assumes that:
2^(p-1)=1(mod p).
Or more generalized:
2^(phi(m))=1(mod m).
By order you mean?:
If e is the smallest exponent such that a^e=1(mod m) and (a,m)=1
Then so, a is of [U]order [/U]e modulo m?[/quote]
So may 2 be a primitive root of p?

[QUOTE=blob100;213537]Ha, I tought you are asking about a set becuase of element is a part of a set, sorry.
Fermat little theorem assumes that:
2^(p-1)=1(mod p).
Or more generalized:
2^(phi(m))=1(mod m).
By order you mean?:
If e is the smallest exponent such that a^e=1(mod m) and (a,m)=1
Then so, a is of [U]order [/U]e modulo m?[/QUOTE]

Correct. What facts do you know about the order of an element?

[QUOTE=blob100;213540]So may 2 be a primitive root of p?[/QUOTE]

Definitely NOT!