Re: Prize Puzzles : Two more Primality Conjectures
- --- In email@example.com, "aldrich617" <aldrich617@...>
> 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)
> 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
> 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.
> 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
So = (2*n+1) + ... + 3*n
= (3*n+1)+3*n/2 - (2*n+1)*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
Inserting the formula for Se gives
2*M-1+p = (5*n^2-n)/p
(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)
f1 = 5*n^2-p^2
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^2So, however the factors of (4*p^2) divide f1*f2, the result must be a
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).
- --- On Mon, 10/6/08, Mike Oakes <mikeoakes2@...> wrote:
> --- In firstname.lastname@example.org, "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.