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Re: polynomials can't give only primes  sci.math #54025
Posted:
Sep 6, 1996 1:19 PM


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+jq]^2 + [(gk+2g+k+1)(h+j)+hz]^2 + : [16(k+1)^3 (k+2) (n+1)^2 +1f^2]^2 + [ 2n+p+q+ze ]^2 + : [ e^3 (e+2)(a+1)^2 + 1  o^2]^2 + [(a^21)y^2 + 1  x^2]^2 + : [16r^2 y^4 (a^21) + 1u^2]^2 + : [ ( (a+u^2 (u^2a))^2  1 ) (n+4dy)^2 + 1  (x+cu)^2]^2 + : [(a^21)l^2 + 1  m^2]^2 + [ai+k+1li]^2 + [n+l+vy]^2 + : [p+l(an1)+b(2an+2an^22n2)m]^2 + : [q+y(ap1)+s(2ap+2ap^22p2)x]^2 + : [z+pl(ap)+t(2app^21)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! (Please don't flame me if this is wrong. I haven't tested it myself, but think I remember having read it somewhere).
Tord Romstad



