## Twin polynomials contest

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• I ask about triple primes (p(d),q(d),r(d)) with r(d)=32d^4+8d^2+1 Does somebody find a solution ???
Message 1 of 19 , Jan 24, 2006

r(d)=32d^4+8d^2+1

Does somebody find a solution ???
• I have done a mistake : of couse Jacques is the leader !
Message 2 of 19 , Jan 24, 2006
I have done a mistake : of couse Jacques is the leader !
• If we are having a contest we need to at least seed it with a titanic prime d=253!+85052 yields a 1001 digit prime 4*((253!+85052)^2)+1 (along with its 1501
Message 3 of 19 , Jan 24, 2006
If we are having a contest we need to at least seed it with a titanic
prime
d=253!+85052 yields a 1001 digit prime
4*((253!+85052)^2)+1
(along with its 1501 digit companion)
4*(253!+85052)*(2*(253!+85052)^2+1)+1
Cheers
Ken
<henridelille@y...> wrote:
>
> Dear all,
>
> I propose the following contest:
>
> Consider these two polynomials:
>
> p(d)=4d^2+1
> q(d)=4d(2d^2+1)+1
>
> The polynomials are often prime for the same parameter d.
> I conjecture that there are an infinty of prime pairs (p,q).
>
> My own record is: d=502 -> p=1008017 and q=1012050073
>
> Contest : find very high parameters d with (p,q) prime pairs
>
• Wonderfull Ken ! I follow up my explanation : Theorem (Fermat) - Every prime p=4k+1 write p=a^2+b^2 where a,b are natural numbers. Theorem - A prime number is
Message 4 of 19 , Jan 24, 2006
Wonderfull Ken !

I follow up my explanation :

Theorem (Fermat) - Every prime p=4k+1 write p=a^2+b^2 where a,b are
natural numbers.

Theorem - A prime number is well approximated by the square of
rationnal numbers p=(u/v)^2+e/v^2 and e small, v as large as we want.
(it is related to continued fractions)

Then a prime 4k+1 is related to a quasi "congruent number" d=ab/2 (see
previous message) in the sense:

p=a^2+b^2=(u/v)^2+e/v^2 and e is small compared to v.

to be continued...
• ... Let x=2d. Then the goal is primes x^2+1 and x^3+2x+1. x^2+1 would be a generalized Fermat prime and many of them are known. I got no hit on x^3+2x+1 from
Message 5 of 19 , Jan 24, 2006
henridelille wrote:

> p(d)=4d^2+1
> q(d)=4d(2d^2+1)+1
>
> Contest : find very high parameters d with (p,q) prime pairs

Let x=2d. Then the goal is primes x^2+1 and x^3+2x+1.
x^2+1 would be a generalized Fermat prime and many of them are known.
I got no hit on x^3+2x+1 from the tested GF primes in
The Prime Pages database.

small GF primes.
b^1024+1 and b^2048+1 gave nothing.
b^512+1 gave several hits corresponding to these x:
65710^256
177536^256
248578^256
1072214^256
4561880^256

PrimeForm/GW has proven all primes.

--
Jens Kruse Andersen
• ... 4*(2^1024+73006634)^2+1 8*(2^1024+73006634)^3+4*(2^1024+73006634)+1 32*(2^1024+73006634)^4+8*(2^1024+73006634)^2+1 4*(2^1024+403799789)^2+1
Message 6 of 19 , Jan 25, 2006
<henridelille@y...> wrote:
>
>
> r(d)=32d^4+8d^2+1
>
> Does somebody find a solution ???

4*(2^1024+73006634)^2+1
8*(2^1024+73006634)^3+4*(2^1024+73006634)+1
32*(2^1024+73006634)^4+8*(2^1024+73006634)^2+1

4*(2^1024+403799789)^2+1
8*(2^1024+403799789)^3+4*(2^1024+403799789)+1
32*(2^1024+403799789)^4+8*(2^1024+403799789)^2+1

4*(2^1024+693904189)^2+1
8*(2^1024+693904189)^3+4*(2^1024+693904189)+1
32*(2^1024+693904189)^4+8*(2^1024+693904189)^2+1

The largest of each triplet is titanic. 10 GHz hours, but 3 was twice
the yield I expected in that time. Assuming O(D^5) growth, a
1000-digit smallest prime should be doable in <1GHz week. I'll not do
that myself, but can donate sieve-technology to anyone who wants to.
(My guess is that PFGW -f would be less than half speed.)

Can we have a complete explanation now both the pairs and the triplets

Phil
• I am impressed by your all results. I think yhat everyone has won. I continue my explanation. (see my 2 previous mails) As a prime 4k+1 can be considered
Message 7 of 19 , Jan 25, 2006
I am impressed by your all results. I think yhat everyone has won.

I continue my explanation. (see my 2 previous mails)

As a prime 4k+1 can be considered connected to a "quasi-congruent"
number d=ab/2 it is natural to ask whether there is a elliptic curve
attached y^2=x^3-d^2x.

In fact as this not really a congruent number the elliptic curve
shall be modified. For some heuristic reasons there are two families
which hit : E1 y^2=8u(x^3-d^2x)+v and E2 y^2=8u(x^3+d^2x)-v, the
parameters u and v depending of the prime p.

Now, in the particular case of primes p(d)=4d^2+1, the parameters
v=1 is good for both E1 and E2 but u for E2 is u=1 and for E1 is u=
(2e)^2 where e is a natural number depending of the continued
fraction of sqrt(p). It is too long to explain the details...

Now put x=p/2 in E1 and E2 and you get:

E1 z1=32d^4+8d^2+1
E2 z2=4d(2d^2+1)

Here is our polynomials:

p(d)=4d^2+1
q(d)=z2+1=4d(2d^2+1)+1
r(d)=z1=32d^4+8d^2+1

I have immediately remarked that for small parameters d the numbers
p,q,r are often prime together.

I don't know exactly why but now you know that it is connected to:
- congruent numbers
- continued fractions
- elliptic curves with rational solutions

It requests further deep investigations to undersand this phenomena.

You can find an more detailed overview (in french) on my website

Note number 4

Enjoy primes
• Hi, My final entry in contest d=450!+42455 4*((450!+42455)^2)+1 (digits:2002) 4*(450!+42455)*(2*(450!+42455)^2+1)+1 (digits:3002) Both certified prime using
Message 8 of 19 , Jan 25, 2006
Hi,
My final entry in contest

d=450!+42455
4*((450!+42455)^2)+1 (digits:2002)
4*(450!+42455)*(2*(450!+42455)^2+1)+1 (digits:3002)

Martin).
The first taking 1h 18mn 22s the second 8h 5mn 2s

Primes found in a couple of hours using a pfgw script file
(source available on request)
Cheers
Ken
<henridelille@y...> wrote:
>
> I am impressed by your all results. I think yhat everyone has won.
>
> I continue my explanation. (see my 2 previous mails)
>
> As a prime 4k+1 can be considered connected to a "quasi-congruent"
> number d=ab/2 it is natural to ask whether there is a elliptic
curve
> attached y^2=x^3-d^2x.
>
> In fact as this not really a congruent number the elliptic curve
> shall be modified. For some heuristic reasons there are two
families
> which hit : E1 y^2=8u(x^3-d^2x)+v and E2 y^2=8u(x^3+d^2x)-v, the
> parameters u and v depending of the prime p.
>
> Now, in the particular case of primes p(d)=4d^2+1, the parameters
> v=1 is good for both E1 and E2 but u for E2 is u=1 and for E1 is u=
> (2e)^2 where e is a natural number depending of the continued
> fraction of sqrt(p). It is too long to explain the details...
>
> Now put x=p/2 in E1 and E2 and you get:
>
> E1 z1=32d^4+8d^2+1
> E2 z2=4d(2d^2+1)
>
> Here is our polynomials:
>
> p(d)=4d^2+1
> q(d)=z2+1=4d(2d^2+1)+1
> r(d)=z1=32d^4+8d^2+1
>
> I have immediately remarked that for small parameters d the
numbers
> p,q,r are often prime together.
>
> I don't know exactly why but now you know that it is connected to:
> - congruent numbers
> - continued fractions
> - elliptic curves with rational solutions
>
> It requests further deep investigations to undersand this
phenomena.
>
> You can find an more detailed overview (in french) on my website
>
> Note number 4
>
> Enjoy primes
>
• Jens is the winner for the twin contest and Phil wins the triple contest with impressive solutions. I hope that you are not deceived by my explanations. An
Message 9 of 19 , Jan 26, 2006
Jens is the winner for the twin contest and Phil wins the triple
contest with impressive solutions.

I hope that you are not deceived by my explanations.

An interesting point is to look at randomly choosen twin or triple
polynomials and compare their performance in giving primes to our
p,q,r.
• An add-on about my explanations: If someone is able to prove that for all primes there is a set of parameters u,v in order that p/2 belongs to the elliptic
Message 10 of 19 , Jan 26, 2006

If someone is able to prove that for all primes there is a set of
parameters u,v in order that p/2 belongs to the elliptic curves E1 or
E2, then he would prove the longstanding problem : there is an
infinity of primes x^2+1
• Hello henridelille, I think it was Euler, not Fermat, who said p = 4k+1 = a²+b². And this is a quality not only of prime numbers, f.e. 25 or 85. Werner
Message 11 of 19 , Jan 26, 2006
Hello henridelille,

I think it was Euler, not Fermat, who said p = 4k+1 = a²+b². And this
is a quality not only of prime numbers, f.e. 25 or 85.

Werner
• Indded it is Fermat Euler s theorem states p=6n+1 - p=x^2+3y^2
Message 12 of 19 , Jan 26, 2006
Indded it is Fermat

Euler's theorem states p=6n+1 -> p=x^2+3y^2
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