blob100: I would recommend Steven Strogatz's columns.
[url]http://opinionator.blogs.nytimes.com/category/steven-strogatz/[/url]

They are in reverse chronological order - it is a blog after all, so start with the first one.

[QUOTE=blob100;211690]I think I solved problem 2.
We can show k2^(n+2)+1 as 4k2^n+1.
Where k2^(n+1)+1 can be shown as 2k2^n+1.
We can show that every number of the form 2^(2^n) is of the form a^2,
example: 2^(2^5)=65536^2.
so becuase of 2^(2^n)-1 factors are equivallent to the factors of a^2 which are 4k+1, we can say easily that 2^(2^n)+1's factors are not the same as the factors of 2^(2^n)-1, so we just agree it is of another form (on this situation k2^(n+2)+1).[/QUOTE]

No. This is not even wrong. It isn't a real argument at all.

[QUOTE=R.D. Silverman;211720]No. This is not even wrong. It isn't a real argument at all.[/QUOTE]

I will offer a hint to problem 2. 2^(2^n) + 1 equals 1 mod 8 for n > 1.

[quote]Here are 5 problems from elementary mathematics. Three involve 1st
semester number theory. Show us that you can do these problems.

(1) Consider a disc. (circle plus its interior). It is intersected by
N lines, each line intersecting at two distinct points (i.e. no tangents)
As a function of N, what is the largest number of regions into which the
disc can be divided? Prove your answer by induction.

(2) Let p be prime, and (a,p) = 1. It is well known that a factor of
a^p - 1, other than (a-1) must be of the form 2kp+1. However, for
Fermat numbers, a factor of 2^2^n+1 must equal k * 2^(n+2) + 1.
Explain why the exponent is (n+2) and not (n+1)

(3) Let p be an odd prime and consider the units of Z/pZ. What is the
product of all of the units?

Definition: A unit of a ring is an element that has a multiplicative inverse.

(4) Let odd N be the product of k distinct prime factors, each dividing exactly.
Suppose x^2 = a mod N has at least one solution. How many total solutions are there?

Hint: Consider the Chinese Remainder Theorem

(5) Consider the two vectors V1 = (x1, y1) and V2 = (x2, y2).

A. Prove that x1 x2 + y1 y2 = |V1| * |v2| * cos (theta) where
theta is the angle between the two vectors.

B. Let V3 be the vector obtained by rotating V1 clockwise by angle mu.
Express this vector in terms of (x1, y1)
[/quote]

Knuth labels his problems with a difficulty factor of 0 to 50,
with (from my imperfect memory) 0 = trivially evident, and
50 = unsolved and very difficult.

Would you (RDS) give us your equivalent estimates for these
(and maybe any future) problems?

Here is a problem I made by myself.

denote:
n_1=ln(n)
n_2=ln(ln(n))
...

denote:
0_n=n^0
1_n=n^1
2_n=n^n
3_n=n^(n^n)
...

Here is the problem:
We know that ((i-1)_e)_i=0.
So, I propose:
(c-3)_e<Pi_2(n)<(c-1)_e

Where "c" means: n_c<0, and not complex.
Pi_2(n) is the number of twin prime pairs from n to 0.
This seems true for me.

Here is an example:
For n=100.
c=4.
1_e<Pi_2(100)<3_e.
e<Pi_2(100)<e^(e^e)
e<8<e^(e^e).

I know, the signs are awfull.. sorry.

[QUOTE=blob100;211792]Here is a problem I made by myself.

denote:
n_1=ln(n)
n_2=ln(ln(n))
...

denote:
0_n=n^0
1_n=n^1
2_n=n^n
3_n=n^(n^n)
...

Here is the problem:
We know that ((i-1)_e)_i=0.
So, I propose:
(c-3)_e<Pi_2(n)<(c-1)_e

Where "c" means: n_c<0, and not complex.
Pi_2(n) is the number of twin prime pairs from n to 0.
This seems true for me.

Here is an example:
For n=100.
c=4.
1_e<Pi_2(100)<3_e.
e<Pi_2(100)<e^(e^e)
e<8<e^(e^e).

I know, the signs are awfull.. sorry.[/QUOTE]

STOP!!!!

I give up. You just won't learn, will you?
Stop it with the conjectures, proposals, "seems true to me", and other
nonsense. It is a clear case of trying to run before you can walk.

Go solve the 5 elementary problems that I gave. 3 are elementary number
theory which is not a secondary school topic, but is doable by someone with
a good grasp of secondary school math. The other two are secondary school problems. Show us that you know how to do elementary induction
arguments, understand basic vector algebra, and can do simple number
theory problems. The three number theory problems that I posed can
all be solved from elementary theorems that you should have seen.

Learn how to do simple proofs and derivations from elementary principles.

[quote=R.D. Silverman;211801]STOP!!!!

I give up. You just won't learn, will you?
Stop it with the conjectures, proposals, "seems true to me", and other
nonsense. It is a clear case of trying to run before you can walk.

Go solve the 5 elementary problems that I gave. 3 are elementary number
theory which is not a secondary school topic, but is doable by someone with
a good grasp of secondary school math. The other two are secondary school problems. Show us that you know how to do elementary induction
arguments, understand basic vector algebra, and can do simple number
theory problems. The three number theory problems that I posed can
all be solved from elementary theorems that you should have seen.

Learn how to do simple proofs and derivations from elementary principles.[/quote]
I don't know vectors, and I know induction really basically.. sorry I'm stupid.
I don't know what is a ring, the definition won't help me.
The problem with the fermat number must ask for knowledge.
And I found the function for the disc..
And the problem with the chinese reminder theorem is solved easily by the theorem.
Will you give me other problems?

[QUOTE=blob100;211831]And I found the function for the disc.[/QUOTE]

Please show your work on this one. I suspect that you have the answer but will have trouble showing the proof in a mathematically rigorous manner. We can help with that. Seeing your work will help us pick appropriate additional problems.

f(N)=F(N-1)+N

[QUOTE=blob100;211865]f(N)=F(N-1)+N[/QUOTE]

A promising start. You really should define all the terms - you have used N, f, and F. You could wiggle away on the grounds that they are implicit in the problem statement and a typo to have both f and F, but it's better get in the habit.

The answer is incomplete because there isn't any way to start evaluating the f's (or F's). We usually handle that by giving the value for some small value of N - perhaps 0 or +1 or -1. With that, it's possible to calculate f (for larger integral N).

The proof is incomplete because you haven't offered any rationale for the statement.

And a closed form solution, so that we could calculate f(100000000000) without all the intermediate terms, would be nice if you can find one.

Your next assignment is to see how many of these shortcomings you can fix and post again.

William

[QUOTE=blob100;211831]I don't know vectors, and I know induction really basically.. sorry I'm stupid.

[/QUOTE]

Not stupid. Merely uneducated.

[QUOTE]

I don't know what is a ring, the definition won't help me.
[/QUOTE]

OK. There are a number of ways to define a ring. For the moment,
instead of 'ring', replace it with 'integers mod N, for N > 1'. The
units are elements of the reduced residue class mod N that have a
multiplicative inverse mod N. What is the product of all of the units mod N?

[QUOTE]


The problem with the fermat number must ask for knowledge.
[/QUOTE]

According to a prior post in which you claimed knowledge of a large
number of things, you have what is needed to solve this problem.
It is only 3 or 4 steps.


[QUOTE]
And I found the function for the disc..
And the problem with the chinese reminder theorem is solved easily by the theorem.
[/QUOTE]

Show your work.

[QUOTE]
Will you give me other problems?[/QUOTE]

I have not seen any of your work on the current problems!

Knowledge of vectors is fundamental to a great many things.
Think of the vector (x1, y1) in the Euclidean plane as the line
segment from (0,0) to (x1, y1). A vector has a length (also known as
its norm) and a direction. The dot product (also known as an inner
product) of two vectors (x1, y1), and (x2, y2) is the scalar x1*x2 + y1*y2.
Your problem is to show that the dot product of (x1, y1), and (x2, y2)
equals the product of their norms times the cosine of the angle between
them.

When I have seen your work on the current set of problems, and after
I have led you through their complete solution, I will present more problems.