# Fermat primes

1. Jul 9, 2009

### chhitiz

for all fermat no.s, fi-1=[fi-1-1]2
all fermat no.s are of form 6Ni+5
Ni+1=6Ni2+8Ni+2
a no. of form 6N+5 is composite only when N is of form 6n1n2+5n1+n2
if we could assume that all fermat no.s after f4 are not prime,
that means all Nis are of form 6n1n2+5n1+n2, i>4
that means that either for any N if N is of form 6n1n2+5n1+n2, Ni+k is of same form, and the no.s in between are coincidentally, of the same form
OR
if f(Ni)=6Ni2+8Ni+2
some fk(Ni) is of form 6n1n2+5n1+n2
if either of these is proved to be true, it could be proved that no. of fermat primes are finite. am i making any sense at all?

2. Jul 13, 2009

### ramsey2879

35 is composite and is of the form 6n+5. I had trouble getting it in the form 6*A*B +5*A + B but then found A=7, B = 0 fits as I was writing this. So my mistake to only quote that portion of your post.
6*0+5 = 5 = 2^2+1 = Fermat Prime 1 so $$N_{1} = 0$$

6*0*0 +8*0 +2 = 2 so $$N_{2} = 2$$ check $$6*2+5 =2^{4} +1 = F_{2}$$

6*2*2 +8*2 +2 = 42 so $$N_{3} = 42$$ check $$6*42+5 =2^{8} +1 = F_{3}$$

6*42*42 +8*42 +2 = 2 so $$N_{4} = 10922$$ check $$6*2+5 =2^{2^16} +1 = F_{4}$$

Continuing

$$N_{5} = 715827882$$ check $$6*N_{5}+5 =2^{32} +1 = F_{5}$$

What you are saying is that since the number of Fermat primes is known to be finite, then the Fermat numbers after the largest Fermat prime are of the form $$6_{N}+5$$
where N is $$6*(N_{i})^{2} + 8*N_{i} +2$$ where $$N_{i}$$ is the N of the prior Fermat number, and also of the form $$6*A*B + 5*A + B$$. Did I understand you correctly?

3. Jul 13, 2009

### Mensanator

4. Jul 13, 2009

### ramsey2879

I looked at this again and it doesn't make any sense since all integers N are of the form 6AB+5A+B. Just make either A or B zero. For instance N=12 is composite A = 0 B = 77 is the only solution of form 6AB + 5A + B. For some N, 6N+5 is prime; for other N, 6N+5 is composite.

5. Jul 18, 2009

### chhitiz

yes you did. also, n1 is always >0 and n2 is >=0

6. Jul 19, 2009

### ramsey2879

That statement is belied by the fact that there is no way to put 6*12+5 into the form without making n1 ZERO; yet 77 is clearly composite. So something is still amiss about your post here. Now if you could have shown that all numbers that are both of the form 6AB + 5A + B with A > 0 and of the form 6N + 5 are composite then I would have found something more than just banter. However, you can't for when A=1 and B = 6 we get a prime number which is 6*7 + 5. Therefore it is not enough to show that numbers of the form 6Ni + 5 are also of the form 6AB + 5A + B to show that such numbers are composite.

7. Jul 20, 2009

### chhitiz

oh i am sorry, it's not 6n+5 that is in form of 6ab+5a+b.
the n is of that form.

8. Jul 20, 2009

### ramsey2879

My mistake

Now I see what you are saying

all numbers 6*N + 5, when n is of the form 6ab + 5a + b, can be factored as (6a+1)*(6b+5) and are composite when neither term = 1.

To show that N is of the first form is in effect the same as factoring and would be a difficult problem.

The only salvation is that indeed if you can show that the N is of the appropriate form and neither of the factors is 1 then you obviously have a composite number.

Last edited: Jul 20, 2009
9. Jul 20, 2009

### chhitiz

yes. now i already tried all possible ways to express Ni+1 as 6n1n2+5n1+n2 for every Ni of the same form, and that is not possible. but it is still possible for some Ni+k. if we can show that, then we can show that fermat primes are finite.

10. Jul 21, 2009

### ramsey2879

Because $$F(i)$$ is prime for i = 1,2,3,4, it would not be surprising if k would have to be greater than 2. k=3 is possible since one factor could be 1 when F(4) is determined. However, it may be that
$$6N_{i}^2 + 8N_{i} + 2$$ no matter how many k times reiterated to form subsequent N could never be expressed by different terms using $$N_i$$ to determine the A and B of the form (6A+1)*(6B+5) for the value of F(i+k)).

11. Jul 22, 2009

### chhitiz

yes i know. i just wish someone with enough resources and intelligence can determine what i can't. that is why i posted this.

12. Jul 24, 2009

### chhitiz

is what i am saying highly improbable? somebody please comment.

13. Jul 30, 2009

### ramsey2879

Have you tried substituting 6AB + 5A + B for $$N_{i}$$?

14. Aug 1, 2009

### chhitiz

yes i have. but that is for the first of two possible scenarios. in what you have quoted i am implying that fk(Ni) or Ni+k as you would better have it, itself happens to be in the form of 6ab+5a+b. and in this case obviously, k>=5. that makes an expression of power 32. and to express even the 'k=5 expression' as 6ab+5a+b, i need to solve 17 equations with very large coefficients all of which must happen to have corresponding solutions.
as regards the first scenario, i've already checked Ni+1 and it doesn't hold.

15. Aug 1, 2009

### ramsey2879

At first glance I think you may not need to worry about powers of 32 to start

A=0, B = 2 gives Ni for 2^4 +1

The largert power for 2^32 + 1 is 6^7(AB)^8 and because A = 0 There are only 9 terms to play with
N4 = B
N8 = 6*B^2+8*B+ 2
N16 = 6*(6*B^2+8*B+2)^2 +8*(6*B^2+8*B+2) +2
N32 = 6*(...)^2 +8*(...) + 2

If you can put N32 into the form 6ab + 5a + b, where a>0 that would give you a head start for the more general form with A > 0. Note also that this is using k = 3. Of course you may have already have tried this.

16. Aug 2, 2009

### ramsey2879

I have been thinking about this and want you to know that I am very uneducated in this subject. It seems that you have spent a lot of time and no one else has responded to this thread even though you pleaded for guidance. The factors for $$N_{i}$$ for small i are well known and most likely you are not the first one to try this approach. Therefore in spite of your assiduous efforts, I believe the lack of response does not bode well for success. My own naive suggestion is just too good to be correct for if correct then there would be no need to solve for the case A>0 and the fairly uniform pattern of the factors would have been recognised before.

Last edited: Aug 2, 2009
17. Aug 2, 2009

### chhitiz

if i understand you correctly yes i have tried your suggestion. and i think we will have to start with N1 as it it self is of form 6x2+8x+2. the problem still happens to be an expression of power 32.
btw i read your blog on fermat's last theorem and really would like u to complete it. i really don't know much about the proof.

18. Aug 3, 2009

### ramsey2879

Yes B = N1. When k = 4, I found

$$N_{5} = 6^{15}B^{16} + ... +17179869184B + 715827882$$ If there is a solution for a and b of the equation $$N_{5} = 6ab + 5a + b$$, a > 0 , I found that
$$a = C_{1}B^{i} + ... +4264852B + 2156$$ and
$$b = C_{2}B^{16-i}+..+1116736B + 106$$. That is as far as I got. Still I decided that even though I got that far, that it is not likely to have a solution, just as the case for k = 3 had no solution. and the power 6^15 is greater than 2^38

19. Aug 4, 2009

### chhitiz

you mean 2^32.
if that is the case then i guess we need to look for higher terms(as power of 2 is increasing exponentially) anyways it would be better to run this through a mainframe computer or something till we get our required value of 'k'.

20. Aug 5, 2009

### ramsey2879

I was trying to say that even with k = 4 you have a conumdrum of a problem to solve, and I don't think it likely that there is a solution.

BTW, the form 6N^2 + 8N + 2 gives Ni for squares of 6N + 4 plus 1

That is 6(6N^2 + 8N + 2) + 5 = (6N+4)^2 + 1 regardless of the N and if you are correct if Ni+k can be put into the form 6AB + 5A + B (A>0) then

$$(6N+4)^{2}^{i+k}+1$$ would have to be composite for all integer N and I don't think that would be so.