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Topic: polynomials can't give only primes - sci.math #54025
Replies: 5   Last Post: Sep 13, 1996 5:00 PM

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Robert Israel

Posts: 11,902
Registered: 12/6/04
Re: polynomials can't give only primes - sci.math #54025
Posted: Sep 6, 1996 7:55 PM
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In article <50pmfp$aa2@beyla.ifi.uio.no>, tordro@ifi.uio.no (Tord Kallqvist Romstad) writes:
|> Richard Pinch (rgep@dpmms.cam.ac.uk) wrote:
|> : In article <50odl2$ecr@nuscc.nus.sg>,
|> : sci50090@leonis.nus.sg (VeLaGaMist) writes:
|> : |> In anycase, it has been proved that there is indeed an integral
|> : |> polynomial that gives primes whenever it takes positive values.
|> : |>
|> : |> I do not have an idea of the explicit form of the polynomial. Could
|> : |> anyone enlighten me on this?
|>
|> : Accoring to my notes, it is
|>
|> : (k+2){1-([wz+h+j-q]^2 + [(gk+2g+k+1)(h+j)+h-z]^2 +
|> : [16(k+1)^3 (k+2) (n+1)^2 +1-f^2]^2 + [ 2n+p+q+z-e ]^2 +
|> : [ e^3 (e+2)(a+1)^2 + 1 - o^2]^2 + [(a^2-1)y^2 + 1 - x^2]^2 +
|> : [16r^2 y^4 (a^2-1) + 1-u^2]^2 +
|> : [ ( (a+u^2 (u^2-a))^2 - 1 ) (n+4dy)^2 + 1 - (x+cu)^2]^2 +
|> : [(a^2-1)l^2 + 1 - m^2]^2 + [ai+k+1-l-i]^2 + [n+l+v-y]^2 +
|> : [p+l(a-n-1)+b(2an+2a-n^2-2n-2)-m]^2 +
|> : [q+y(a-p-1)+s(2ap+2a-p^2-2p-2)-x]^2 +
|> : [z+pl(a-p)+t(2ap-p^2-1)-pm]^2
|> : ) }
|>
|> : (the layout may help show why it is not of much practical use!).
|>
|> There is one more reason it is not very practical:
|> Most of the time, the formula just give you the primes 2 and 3!

No. Most of the time it gives non-positive values.
Note that the polynomial is of the form
(k+2){ 1 - (sum of squares of quantities in square brackets) }
so the value will be either k+2 (if all the quantities in square brackets
are 0) or something <= 0. And, as it turns out, the only way all
the quantities in square brackets can be 0 is if k+2 is prime.

Robert Israel israel@math.ubc.ca
Department of Mathematics (604) 822-3629
University of British Columbia fax 822-6074
Vancouver, BC, Canada V6T 1Y4







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