## 2a. HARDY LITTLEWOOD CONJECTURE ANALISYS. FIRST PROPOSITION.

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• ... [Meaning if p = 1 mod 6, k = (p-1)/6, and if p = 5 mod 6, (k = (p+1)/6. ... Simple proof: In case p = 6 k + 1, and n = 6 j + 1 pn + k = (6 k + 1) * (6 j +
Message 1 of 1 , Jan 6, 2011
> 2a. HARDY LITTLEWOOD CONJECTURE ANALISYS. FIRST PROPOSITION.
> Posted by: "ALBERTO" arfzelaya@... albrtzlya
> Date: Wed Jan 5, 2011 6:25 pm ((PST))

> Let p any prime of the form 6k+-1

[Meaning if p = 1 mod 6, k = (p-1)/6, and if p = 5 mod 6, (k = (p+1)/6.

> and n any integer number
> and k the k value corresponding to each prime of the form 6k+-1

> (for instance k values corresponding to prime 5 is 1 because 5=(6*1)-1
> and k value for 109 is 18 because 109=6*18+1)

> Therefore the algebraic expression
> pn+-k is equivalent to 6ab+-a+-b

Simple proof:

In case p = 6 k + 1, and n = 6 j + 1
pn + k
= (6 k + 1) * (6 j + 1) + k
36 j k + 6 j + 6 k + 1 + k
= 6 k (6 j + 1) + (6 j + 1) + k ; a = k; b = 6j + 1;

In case p = 6 k + 1, and n = 6j - 1
pn - k
=(6k+1)(6j-1) - k
=36 k j + 6 j - 6 k -1 - k
= 6 k (6 j - 1) + (6 j - 1) - k; a = k; b = 6 j - 1

In case p = 6 k - 1, and n = 6 j + 1
pn - k
= (6 k - 1) (6 j + 1) - k
= 36 k j + 6 k - 6 j - 1 - k
= 6 k (6 j + 1) - (6 j + 1) - k ; a = k; b = (6j+1)

In case p = 6k -1, and n = 6 j - 1
pn + k
= (6k - 1)(6j-1) + k
= 36 kj - 6 k - 6 j + 1 + k
= 6 k (6j - 1) - (6j-1) + k ; a = k, b = (6 j - 1)

> In other words if an integer x is not equal to pn+-k it won't be
> equal to 6ab+-a+-b

Proof of if and only if:

If p = 6 k + 1, and b = n, and a = k,
then
pn + k = (6 k + 1)n + k = (6 a + 1) b + a = 6 a b + b + a

If p = 6 k - 1, and b = n, and a = k,
then
pn - k = (6k-1)n - k = (6 a - 1) b - a = 6 a b - b - a

> I wish to know if in your opinion this proposition is right or wrong.
> Kind regards,

Seems to be an algebraic identity.

> Alberto Zelaya

Kermit Rose
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