[QUOTE=blob100;213136]The solve of problem (1):
Let the formula be:
F(N)=F(N-1)+N.
For N=1, F(1)=F(0)+1=2, which is true.
From now, we try to prove that if the formula is true for N, it is true for N+1.
If we want a maximal areas for N+1 lines, the line number N+1, will be needed to cross the whole other lines and of course second times the circle.
*If we will add a new line, every area it crosses, will devided to two areas.
*If The new line is crossed by two consecutive places (places that between then there is no cross of a line with the line), we say, the new line made a new area (devided an area).
Easily we say there are N+1 consecutive pairs, and so, N+1 new areas.
We say F(N+1)=(F(N)=last areas)+(N+1=new areas).[/QUOTE]

This is handwaving.

You first need to express F(N) as a polynomial. I showed you how
to do this. You then need to prove that the polynomial gives the same
results as the recursive formula. One uses a simple induction for this.

If you do not know what induction is, or how to use it, please say so and
we will help.

Have you made any progress on the other problems? I presented the solution method for one of them (number of solutions to x^2 = a mod N), almost
completely.

I will give a hint to the problem of finding the product of the units mod N:

Pair each element with its inverse.


The problem involving vectors only requires some elementary geometry and
trig. This is one you should be able to do without help, [b]if[/b] you know
the material that you claimed to know.

[quote=R.D. Silverman;213205]This is handwaving.

You first need to express F(N) as a polynomial. I showed you how
to do this. You then need to prove that the polynomial gives the same
results as the recursive formula. One uses a simple induction for this.

If you do not know what induction is, or how to use it, please say so and
we will help.

Have you made any progress on the other problems? I presented the solution method for one of them (number of solutions to x^2 = a mod N), almost
completely.

I will give a hint to the problem of finding the product of the units mod N:

Pair each element with its inverse.


The problem involving vectors only requires some elementary geometry and
trig. This is one you should be able to do without help, [B]if[/B] you know
the material that you claimed to know.[/quote]
I think I know what induction is.
Induction is (how i think):
A way to prove a proposition which propose about the whole natural numbers.
The way of proving by induction is:
Showing that for N=1 the proposition is true,
And showing that if it is true for N, it is true for N+1.

I have a question:
For a^p-1 the factors are 2kp+1 right?
If I have a^n-1 for integer n, will the factors be different then 2kn+1, can they be 2kn+1, or these cannot be? (this is usefull for one of your problems).
If for a^n-1, the factors are not 2kn+1, I have an error with the factors problem:
We can see that (2^(2^m)+1)(2^(2^m)-1)=(2^(2^(m+1))-1).
So we can prove that 2^(2^(m+1))-1 is able to be devided by
k2^(m+2)+1, where 2^(2^m)+1 is not devided by this (which means,
2^(2^m)+1 is devided by this).
We see that the factors of 2^(2^(m+1))-1 are not of the form 2kn+1.
But we see the n is 2^(m+1), so we have 2k2^(m+1)+1=k2^(m+2)+1..
Seems wierd..
But let's continue:
The factors of 2^(2^(m))-1 are not 2k2^m+1=k2^(m+1)-1, becuase of n=m.

[quote=R.D. Silverman;213205]This is handwaving.

You first need to express F(N) as a polynomial. I showed you how
to do this. You then need to prove that the polynomial gives the same
results as the recursive formula. One uses a simple induction for this.

If you do not know what induction is, or how to use it, please say so and
we will help.

Have you made any progress on the other problems? I presented the solution method for one of them (number of solutions to x^2 = a mod N), almost
completely.

I will give a hint to the problem of finding the product of the units mod N:

Pair each element with its inverse.


The problem involving vectors only requires some elementary geometry and
trig. This is one you should be able to do without help, [B]if[/B] you know
the material that you claimed to know.[/quote]

F(N)=N^2-((N^2-N)/2)+1=((N^2+N+2)/2)

[QUOTE=blob100;213236]F(N)=N^2-((N^2-N)/2)+1=((N^2+N+2)/2)[/QUOTE]

This is the correct polynomial. Now, try proving it by induction.
Show that it satisfies F(N) = N + F(N-1)

[QUOTE=blob100;213229]I have a question:
For a^p-1 the factors are 2kp+1 right?
If I have a^n-1 for integer n, will the factors be different then 2kn+1, can they be 2kn+1, or these cannot be? (this is usefull for one of your problems).
If for a^n-1, the factors are not 2kn+1, I have an error with the factors problem:
We can see that (2^(2^m)+1)(2^(2^m)-1)=(2^(2^(m+1))-1).
So we can prove that 2^(2^(m+1))-1 is able to be devided by
k2^(m+2)+1, where 2^(2^m)+1 is not devided by this (which means,
2^(2^m)+1 is devided by this).
We see that the factors of 2^(2^(m+1))-1 are not of the form 2kn+1.
But we see the n is 2^(m+1), so we have 2k2^(m+1)+1=k2^(m+2)+1..
Seems wierd..
But let's continue:
The factors of 2^(2^(m))-1 are not 2k2^m+1=k2^(m+1)-1, becuase of n=m.[/QUOTE]


STOP!!!

Once again, you are trying to run before you can walk.

If you want to be taken seriously, then show us the work you have
done on the exercizes that I provided.

I have given strong hints on all 3 number theory problems. So far,
we have not seen anything from you.

[quote=R.D. Silverman;213248]This is the correct polynomial. Now, try proving it by induction.
Show that it satisfies F(N) = N + F(N-1)[/quote]
The proposition doesn't need any induction proof, by basic algebra it can be proved well.
F(N)=F(N-1)+N.
F(N)=F(N-2)+N-1+N.
F(N)=F(N-3)+N-1+N-2+N.

F(N)=N^2+1-(S_(U_i)).
Where U_i is the set of the natural numbers to N-1.
F(N)=N^2+1-(1+N-1)(N-1)/2.
F(N)=N^2+1-(N^2-N)/2.
F(N)=(N^2+N+2)/2.

[quote=R.D. Silverman;213320]STOP!!!

Once again, you are trying to run before you can walk.

If you want to be taken seriously, then show us the work you have
done on the exercizes that I provided.

I have given strong hints on all 3 number theory problems. So far,
we have not seen anything from you.[/quote]
You deffiniently missunderstood.
There is no conjecture here, no propositions, and no proofs.
I just asked a question on one of your problems.
I was asked to show why are the factors of fermat numbers are of the form k2^(n+2)+1, instead of k2^(n+1)+1.
And so far, I showed you an error I met.
The question I asked is about the error..

[QUOTE=blob100;213321]The proposition doesn't need any induction proof, by basic algebra it can be proved well.
F(N)=F(N-1)+N.
[/QUOTE]

OK.

[QUOTE]
F(N)=F(N-2)+N-1+N.
F(N)=F(N-3)+N-1+N-2+N.


F(N)=N^2+1-(S_(U_i)).
[/QUOTE]

This last statement is meaningless, since you have not DEFINED S.

[QUOTE]
Where U_i is the set of the natural numbers to N-1.
F(N)=N^2+1-(1+N-1)(N-1)/2.
[/QUOTE]

Under the assumption that S means "sum", you still need to prove,
by induction that sum(i=1, n) i = n(n+1)/2

Once again, the problem is to prove, by induction that if
F(N) = F(N-1) + N, then F(N) = (N^2 + N)/2 + 1.

[QUOTE=blob100;213322]You deffiniently missunderstood.
There is no conjecture here, no propositions, and no proofs.
I just asked a question on one of your problems.
I was asked to show why are the factors of fermat numbers are of the form k2^(n+2)+1, instead of k2^(n+1)+1.
And so far, I showed you an error I met.
The question I asked is about the error..[/QUOTE]

Use the hint that I gave. You have not used it.


F_n = 2^2^n + 1, and if p | F_n, first show that p equals
1 mod 8 for n >= 2. Now ask: what relation is known between the prime 2,
and primes that are 1 mod 8??? From this relation you can then apply a
congruence from Euler.

This is an exercize that can be done in 3 or 4 lines.

[quote=R.D. Silverman;213323]OK.



This last statement is meaningless, since you have not DEFINED S.



Under the assumption that S means "sum", you still need to prove,
by induction that sum(i=1, n) i = n(n+1)/2

Once again, the problem is to prove, by induction that if
F(N) = F(N-1) + N, then F(N) = (N^2 + N)/2 + 1.[/quote]
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.
Is my proof correct?
To prove it by induction too right?