"Dr. Fairman" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Dear Stuart,
> I promised to perform only one question for free and you put two.
> Below is Q2 solution. If you need Q1 solution, contact me privately
> and we shall negotiate on my fee rate.
> You need me to prove that a+b= 2*k (1) Where: a,b any primes and k is
> some natural integer
> Note, that: a) even and odd natural numbers alternate, i.e.
> adjacent difference (as absolute value) is equal 1; b)prime number is
> always odd number (by its definition);
> So, any prime could be represented as (2*n+1) where n is some (not
> any) natural integer number.
> Hence (1) could be rewritten as a/2+b/2=k or
> (2*n+1)/2+(2*m+1)/2=n+m+1 = k
> since n and m are natural integers, it is obvious that k is also
> natural integer, that was to be proved.
Well no I am afraid not, because although for all p prime p = 2*n+1 is true
it is not true that for all n n in N 2*n+1 is prime which is what you would
need for your proof to be valid.
Are you pulling my leg in return? if so touche :-)
If you are not pulling my leg, I would say that the probability that you
have a PhD in mathematics and do not recognise Q2 is vanishingly small.
PS if you can solve Q1 you could make much more money by publshing the
solution in a book.
--
Stuart Gall
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