- I offer a $50 prize to the first person who can submit a verifiable

counterexample or proof by 11/1/8 for either conjecture 1) or 2)

below:

x,A(x),K,M,S,n,p : integers;

Let A(x) = 5x^2 - 5x +1

with K > 3,

and n or p denoting the number of terms in a sequence.

For any value of K first find the sum of the consecutive integers

K1 + K2.....+ Kn = S such that Kn < (3K -1)/2.

Example: if K = 9 then S = 9 + 10 + 11 + 12 = 42, and n = 4.

(note: 40*S + 1 is always a square)

Using the same value of S find any shorter runs of consecutive

integers

M1 + M2.....+ Mp = S where p < n, and M1 > K1.

Example: For S = 42 from above it is evident that 13 + 14 + 15

also equals 42, with M1 = 13 and p = 3.

Conjectures:

1) If any value for S or M1 found by the process described above is

substituted for x in A(x) = 5x^2 - 5x +1 then

A(x) is always composite .

Examples: if S = 42 then A(42) = 8611 = 79*109.

if M1 = 13 then A(13) = 781 = 11 * 71.

2) If all possible composites from 1) above are removed from the

sequence

5x^2 - 5x +1 , then all the remaining values of A(x) are prime.

Aldrich Stevens

ps.

I have it on good authority that the puzzle reprinted below

from 9/8/8 is probably true but very difficult to prove, so

I am extending the contest deadline until 11/1/8.

Prize Puzzle : Primality Conjecture

I offer a $50 prize to the first person who can submit a verifiable

counterexample or proof for the following primality conjecture:

x, A(x), B(x), k, T(k) : integers;

Let A(x) = 5x^2 - 5x +1;

Let B(x) = 25x^2 - 40x + 16;

Let T(k) = 5*(2*k -1)^2 - 4*A(x) ;

If x > 3 and k > x , then A(x) will be prime if no value

T(k) exists such that 1 < T(k) < B(x) is a square. - --- In primenumbers@yahoogroups.com, "aldrich617" <aldrich617@...>

wrote:>

You don't say so, but I assume you are requiring K1 = K ?

> For any value of K first find the sum of the consecutive integers

> K1 + K2.....+ Kn = S such that Kn < (3K -1)/2.

> Example: if K = 9 then S = 9 + 10 + 11 + 12 = 42, and n = 4.

> (note: 40*S + 1 is always a square)

-Mike Oakes - --- In primenumbers@yahoogroups.com, "aldrich617" <aldrich617@...>

wrote:>

Thanks for presenting this extraordinarily interesting challenge.

> I offer a $50 prize to the first person who can submit a verifiable

> counterexample or proof by 11/1/8 for either conjecture 1) or 2)

> below:

>

> x,A(x),K,M,S,n,p : integers;

>

> Let A(x) = 5x^2 - 5x +1

> with K > 3,

> and n or p denoting the number of terms in a sequence.

>

> For any value of K first find the sum of the consecutive integers

> K1 + K2.....+ Kn = S such that Kn < (3K -1)/2.

> Example: if K = 9 then S = 9 + 10 + 11 + 12 = 42, and n = 4.

> (note: 40*S + 1 is always a square)

> Using the same value of S find any shorter runs of consecutive

> integers

> M1 + M2.....+ Mp = S where p < n, and M1 > K1.

> Example: For S = 42 from above it is evident that 13 + 14 + 15

> also equals 42, with M1 = 13 and p = 3.

>

> Conjectures:

>

> 1) If any value for S or M1 found by the process described above is

> substituted for x in A(x) = 5x^2 - 5x +1 then

> A(x) is always composite .

> Examples: if S = 42 then A(42) = 8611 = 79*109.

> if M1 = 13 then A(13) = 781 = 11 * 71.

>

> 2) If all possible composites from 1) above are removed from the

> sequence

> 5x^2 - 5x +1 , then all the remaining values of A(x) are prime.

>

> Aldrich Stevens

>

[And you confirmed offline that indeed K1=K.]

After spending most of the last 2 days on it, I have just proved

conjecture 1).

I don't know if other people are working on it, in which case it

would spoil their fun to present the proof; but if no-one objects,

then I will publish it here in 24 hours' time.

-Mike Oakes - --- In primenumbers@yahoogroups.com, "aldrich617" <aldrich617@...>

wrote:>

[snip]

> I offer a $50 prize to the first person who can submit a verifiable

> counterexample or proof by 11/1/8 for either conjecture 1) or 2)

> below:

>

> x,A(x),K,M,S,n,p : integers;

>

> Let A(x) = 5x^2 - 5x +1

> with K > 3,

> and n or p denoting the number of terms in a sequence.

>

> For any value of K first find the sum of the consecutive integers

> K1 + K2.....+ Kn = S such that Kn < (3K -1)/2.

> Example: if K = 9 then S = 9 + 10 + 11 + 12 = 42, and n = 4.

> (note: 40*S + 1 is always a square)

> Using the same value of S find any shorter runs of consecutive

> integers

> M1 + M2.....+ Mp = S where p < n, and M1 > K1.

> Example: For S = 42 from above it is evident that 13 + 14 + 15

> also equals 42, with M1 = 13 and p = 3.

>

> Conjectures:

>

> 1) If any value for S or M1 found by the process described above is

> substituted for x in A(x) = 5x^2 - 5x +1 then

> A(x) is always composite .

> Examples: if S = 42 then A(42) = 8611 = 79*109.

> if M1 = 13 then A(13) = 781 = 11 * 71.

>

Proof of Conjecture 1):

If K is even, n = K/2.

If K is odd, n = (K-1)/2.

Let's work in terms of n.

Given n, denote the even-valued K by

Ke = 2*n,

and the odd-valued K by

Ko = 2*n+1.

Denote the corresponding values of S by:

Se = (2*n) + ... + (3*n-1)

= (3*n-1)+3*n/2 - (2*n-1)*2*n/2

= (5*n^2-n)/2

and

So = (2*n+1) + ... + 3*n

= (3*n+1)+3*n/2 - (2*n+1)*2*n/2

= (5*n^2+n)/2

Let 1 <= p < n, and suppose that

Se = M + ... + (M+p-1)

= (M+p-1)*(M+p)/2 - (M-1)*M/2

= (2*M-1+p)*p/2

Inserting the formula for Se gives

2*M-1+p = (5*n^2-n)/p

i.e.

(2*M-1) = (5*n^2-n-p^2)/p

Iff the RHS is an odd integer, this gives a solution.

For any such solution we find

A(M) = 5*M^2-5*M+1

= (5/4)*(4*M^2-4*M) + 1

= (5/4)*(2*M-1)^2 - 1/4

= 5*(5*n^2-n-p^2)^2/(4*p^2) - 1/4

= [125*n^4 - 50*n^3 + (5-50*p^2)*n^2 + 10*p^2*n +(5*p^4-p^2)]/(4*p^2)

= f1 * f2 / (4*p^2)

where

f1 = 5*n^2-p^2

and

f2 = 25*n^2-10*n+1-5*p^2

[Remark: this surprising algebraic factorisation of a quartic

polynomial into a product of 2 quadratics is the core of the proof,

and was found by examining a lot of special cases with, in

particular, p=2, 3 and n.]

Now n > p, so

f1 > (4*p^2)

and n >= (p+1), so

f2 = (5*n-1)^2 - 5*p^2>= (5*p+4)^2 - 5*p^2

So, however the factors of (4*p^2) divide f1*f2, the result must be a

> (4*p^2)

product of 2 integer factors each > 1.

Therefore, A(M) is composite, which was to be proved.

The limiting case p = 1 gives M = Se,

so A(Se) is composite, which was also to be proved.

If, instead, we choose the "odd-K" case (Ko and So), then all the

above goes through with just the simple change n => -n.

This concludes the proof of conjecture 1).

-Mike Oakes - --- On Mon, 10/6/08, Mike Oakes <mikeoakes2@...> wrote:
> --- In primenumbers@yahoogroups.com, "aldrich617" wrote:

[snip]

> > Let A(x) = 5x^2 - 5x +1

> > with K > 3,

> > and n or p denoting the number of terms in a sequence.

> >

> > For any value of K first find the sum of the

> consecutive integers

> > K1 + K2.....+ Kn = S such that Kn < (3K -1)/2.

> Proof of Conjecture 1):

Nowhere is it specified that n is maximal. Of course, it's underspecified if you don't assume that, but all that means is that the problem's underspecified.

>

> If K is even, n = K/2.

> If K is odd, n = (K-1)/2.

Phil