## More on the new conjecture with a conclusion if the conjecture is true

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• This conjecture if true will be able to determine whether any number 1mod10 or (-1)mod10 is prime or not. Let a and b be two relatively prime numbers called a
Message 1 of 1 , Jul 2 2:51 AM
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This conjecture if true will be able to determine whether any number
1mod10 or (-1)mod10 is prime or not.

Let a and b be two relatively prime numbers called a primitive pair a,b.
f(a,b) = a^2 - b^2 - a*b which implies that
f(a+b,a) = -a^2 + b^2 + a*b (by substitution)and that
f(2a+b,a+b) = a^2 - b^2 - a*b the original f(a,b)
(The family a+b,a 2a+b,a+b etc are called resultant pairs)

The conjecture is that:
(i) f(a,b) = 1mod10 or (-1)mod 10 or 0mod5

(ii) f(a,b) will generate every prime number of the form (+/-1)mod10 and the
prime number 5 itself

(iii) f(a,b) also generates all the possible composites of the prime numbers
contained in the above set (in which case there is a primitive c,d not =
primitive a,b nor a resultant pair arising from a,b) except that if f(a,b)
is 0mod5, there will be only be an extra primitive pair if the composite
contains 2 factors other than 5

(iv) f(a,b) will never be a prime of form 3mod10 or (-3)mod10 nor will it,
if composite, contain such a prime as a factor.

Thus if a number X = 1mod10 or (-1)mod10 for which no primitive pair can be
found, X is a composite containing an even number of primes of form 3mod10
or(-3)mod10 as factors.

If only one primitive pair exists X is prime.

If more than one primitive pair can be found, X is a composite also (and the
number of such pairs will be found to be a power of 2, which is a new
conjecture).

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