Re: [PrimeNumbers] Re: Complex a*x^n+b*y^n puzzle
----- Original Message -----
Sent: Friday, December 04, 2009 5:31 PM
Subject: [PrimeNumbers] Re: Complex a*x^n+b*y^n puzzle
next conclusion is that any rational prime that can be represented by the sum
of 2 squares (a^2+b^2) will define 4 gaussian primes a+bi,a-bi,b+ai,b-ai.
--- In firstname.lastname@example.org,
"Robdine" <robdine@...> asked:
> Is a gaussian number a+bi prime if a^2+b^2 is a (integer)prime
> and if c^2+d^2 is prime then is also c+di a gaussian prime?
These two question appear to be identical;
the second merely changes the word order
and the labelling in the first, without
altering the mathematics in any respect.
In each case, the answer is yes.
Moreover, if a and b are non-zero rational integers, then
a + I*b is a Gaussian prime if and *only* if
a^2 + b^2 is a rational prime.
Note, however, that 3 + I*0 is a Gaussian prime,
but 3^2 + 0^2 is not a rational prime.
To determine whether z is a Gaussian prime
we may use the following procedure:
1 + I, 2 + I, 3, 3 + 2*I, 4 + I, 5 + 2*I, 5 + 4*I, 6 + I, 6 + 5*I, 7, 7 + 2*I,
[Non-text portions of this message have been removed]
- --- In email@example.com,
"Robdine" <robdine@...> wrote:
> any rational prime that can be represented by the sumThere is only rational prime of the form a^2 + b^2 that yields
> of 2 squares (a^2+b^2) will define 4 gaussian primes
precisely 4 distinct Gaussian primes, namely 2 = 1^2 + 1^2.
If a^2 + b^2 is an odd rational prime, we have
8 [sic] asociates of the Gaussian prime z = a + I*b,
since we may mulitply it and its conjugate z = a - I*b
by the 4 units I, -I, -1, 1, obtaining 8 distinct
Gaussian integers that are prime.