> > Q2 Show that every even number greater than 2 is the sum of two
> > primes
Dr Fairman wrote:
> 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.
Dr Fairman is not answering the question! Q2 states, given some even
number greater than 2, prove that it is made up of the sum of two
primes, not the partial converse which is not only easy, but is what Dr
Fairman is proving, namely: given any two odd primes, then their sum is
even.
Nim.
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