Proving the Finiteness of Fermat Primes

[chhitiz]

for all fermat no.s, f_i-1=[f_i-1-1]²
all fermat no.s are of form 6N_i+5
N_i+1=6N_i²+8N_i+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 N_is are of form 6n₁n₂+5n₁+n₂, i>4
that means that either for any N if N is of form 6n₁n₂+5n₁+n₂, N_i+k is of same form, and the no.s in between are coincidentally, of the same form
OR
if f(N_i)=6N_i²+8N_i+2
some f_k(N_i) 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?

 

[ramsey2879]

for all fermat no.s, f_i-1=[f_i-1-1]²
all fermat no.s are of form 6N_i+5
a no. of form 6N+5 is composite only when N is of form 6n₁n₂+5n₁+n₂

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?

 

[Mensanator]

What about Fermat Prime 0?

 

[ramsey2879]

for all fermat no.s, f_i-1=[f_i-1-1]²
all fermat no.s are of form 6N_i+5
N_i+1=6N_i²+8N_i+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 N_is are of form 6n₁n₂+5n₁+n₂, i>4
that means that either for any N if N is of form 6n₁n₂+5n₁+n₂, N_i+k is of same form, and the no.s in between are coincidentally, of the same form
OR
if f(N_i)=6N_i²+8N_i+2
some f_k(N_i) 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?

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.

 

[chhitiz]

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?

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

 

[ramsey2879]

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

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 6N_i + 5 are also of the form 6AB + 5A + B to show that such numbers are composite.

 

[chhitiz]

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

 

[ramsey2879]

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

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.

 

[chhitiz]

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

 

[ramsey2879]

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

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)).

 

[chhitiz]

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)).

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

 

[chhitiz]

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

 

[ramsey2879]

for all fermat no.s, f_i-1=[f_i-1-1]²
If f(N_i)=6N_i²+8N_i+2
some f_k(N_i) is of form 6n₁n₂+5n₁+n₂

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

 

[chhitiz]

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

yes i have. but that is for the first of two possible scenarios. in what you have quoted i am implying that f_k(N_i) or N_i+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 N_i+1 and it doesn't hold.

 

[ramsey2879]

yes i have. but that is for the first of two possible scenarios. in what you have quoted i am implying that f_k(N_i) or N_i+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 N_i+1 and it doesn't hold.

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.

 

[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.

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.

 

[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.

 

[ramsey2879]

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.

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

 

[chhitiz]

and the power 6^15 is greater than 2^38

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'.

 

[ramsey2879]

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'.

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 N_i 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 N_i+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.

 

[Count Iblis]

http://en.wikipedia.org/wiki/Von_Staudt–Clausen_theorem

p = 2^(2^k) + 1 is a prime if and only if

[B_{2^(2^k)} - B_{2^(2^k - 1)}] Mod 1 not equal to 0

where B_n is the nth Bernoulli number.

 

[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 N_i 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 N_i+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.

I just explored a page on Generalized Fermat Primes and found that the following is a prime

$$167176^{{2})^{15}} + 1$$

Edit I mean raised to the 2^15 th power
See page 831
http://www.ams.org/mcom/2002-71-238/S0025-5718-01-01350-3/S0025-5718-01-01350-3.pdf
since 167176 + 1 = 6*27862 + 5 we now know that k must be greater than 14! and you must deal with powers of 32768. You would need a supercomputer to solve this!

 

[chhitiz]

I just explored a page on Generalized Fermat Primes and found that the following is a prime

$$167176^{{2})^{15}} + 1$$

Edit I mean raised to the 2^15 th power
See page 831
http://www.ams.org/mcom/2002-71-238/S0025-5718-01-01350-3/S0025-5718-01-01350-3.pdf
since 167176 + 1 = 6*27862 + 5 we now know that k must be greater than 14! and you must deal with powers of 32768. You would need a supercomputer to solve this!

you mean to say that (6n+4)^2i+k has to be composite for all n, k has to be greater
than 14. but i don't understand how you got 14. 2¹⁵should not be equal to 2i+k, whatever the value of k we get(which is very high), we can always get a smaller k to satisfy the above eqn.

 

[ramsey2879]

you mean to say that (6n+4)^2i+k has to be composite for all n, k has to be greater
than 14. but i don't understand how you got 14. 2¹⁵should not be equal to 2i+k, whatever the value of k we get(which is very high), we can always get a smaller k to satisfy the above eqn.

What do you mean we can get smaller k? if (6n+4)^{{2}^15} + 1 is prime, for k smaller than 15, N(i+k) can't be put in the form 6AB + 5A + B for A>0 since we know that there is a
N_i where 6N_i+k + 5 is prime and not composite.

 

[chhitiz]

What do you mean we can get smaller k? if (6n+4)^{{2}^15} + 1 is prime, for k smaller than 15, N(i+k) can't be put in the form 6AB + 5A + B for A>0 since we know that there is a
N_i where 6N_i+k + 5 is prime and not composite.

oh yes you are right. i don't know why i was thinking the part where (6n+4)²+1 was involved it wouldn't be part of this recurrence relation. but i tried to get (6N+4)²+1 for N_i+k and i think it is (6N_i+4)²*2²ⁱ+1 and not (6N_i+4)^2i+k+1

 

[ramsey2879]

oh yes you are right. i don't know why i was thinking the part where (6n+4)²+1 was involved it wouldn't be part of this recurrence relation. but i tried to get (6N+4)²+1 for N_i+k and i think it is (6N_i+4)²*2²ⁱ+1 and not (6N_i+4)^2i+k+1

I believe you mean [tex](6N_i +4)^{2i} + 1[/tex]

 

[srijithju]

Well I really donno if ur heading in the right way .. but very nice thinking ...
If I can be of any help , You can express N(i+1) in terms of N(i) .. ( that is if my calculation is correct ) as follows :

N(i+k) = { (6^(3k)) * ( [ { N(i) + (2/3) } ^ 2 ] ^ {2k} ) } - (2/3)


Now can you somehow use a binomial expansion to get your desired representation .. I donno ... but just give it a try anyway ... sorry that no one's out here to help u more

 

[chhitiz]

I believe you mean [tex](6N_i +4)^{2i} + 1[/tex]

i think it is (2(6n+4))²ⁱ+1. not that it matters. it seems i do need a supercomputer. can anybody lend me theirs?