## Re: a*x^n+b*y^n puzzle

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• Hi Mike, What about : a=140 x=2 b=47 y=3 ? jl
Message 1 of 37 , Nov 27, 2009
Hi Mike,

a=140
x=2
b=47
y=3

?

jl

--- In primenumbers@yahoogroups.com, "mikeoakes2" <mikeoakes2@...> wrote:
>
> Here is a small puzzle:-
>
> Find fixed integers a, b, x, y such that the expression
> a*x^n+b*y^n
> is prime for all n in the range 1 <= n <= n_max,
> where n_max is to be as large as possible.
>
> Hint: n_max=9 is certainly achievable.
>
> -Mike Oakes
>
• ... There is only rational prime of the form a^2 + b^2 that yields precisely 4 distinct Gaussian primes, namely 2 = 1^2 + 1^2. If a^2 + b^2 is an odd rational
Message 37 of 37 , Dec 5 7:49 AM
"Robdine" <robdine@...> wrote:

> any rational prime that can be represented by the sum
> of 2 squares (a^2+b^2) will define 4 gaussian primes

There is only rational prime of the form a^2 + b^2 that yields
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.

David
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