Prime number generation by Jones, Sato, Wada and Wiens
Alexandre
that it is a polynomial of degree 26 which is not much. I would also
like it if someone could post an article that would at least make a
brief description of the formula. Thank you.
Alexandre
Raymond Manzoni
The abstract of the paper is here :
http://www.math.ucalgary.ca/~jpjones/abst1976.htm
The paper :
http://www.stat.ualberta.ca/~wiens/reprints/diophantine.pdf
See to :
http://www.bath.ac.uk/~ensab/Primes/
http://www.math.niu.edu/~rusin/known-math/96/primpoly
Enjoy,
Raymond
Chip Eastham
Ribenboim discusses this and similar formulas in his Little Book of
Big Prime Number Records. If I remember, the polynomial produces only
primes among its positive values, but it also has negative values
which may be composite. Also the polynomial is in more than one
variable. Ribenboim gives then current records for both least degree
or least number of variables, and as you might suspect there is a big
trade-off between these goals.
regards, chip
(substitute icx for bellsouth to reply)
John Robertson
> Alexandre wrote:
> > I am looking for sources that discuss this formula.
> The abstract of the paper is here :
> http://www.math.ucalgary.ca/~jpjones/abst1976.htm
>
> The paper :
> http://www.stat.ualberta.ca/~wiens/reprints/diophantine.pdf
>
> See too :
> http://www.bath.ac.uk/~ensab/Primes/
> http://www.math.niu.edu/~rusin/known-math/96/primpoly
Somewhere in the middle of the polynomial, Ribenboim, The Book of
Prime Number Records, first edition, 1988, page 148, has `x-cu', where
Jones, Sato, Wada, and Wiens in their American Math Monthly article,
1976, page 449, have `x+cu'. Does anyone know whether Ribenboim's
`x-cu' is a correction or a misprint?
John Robertson
Michael Bales
Does anyone have values for the 26 variables that actually produce a
positive prime from this formula? (like for k = 0) I tried to
produce one for a few hours with no success. Is it that the variables
have to be unreasonably huge to do it?
Mike Bales
Phil Carmody
> Somewhere in the middle of the polynomial, Ribenboim, The Book of
> Prime Number Records, first edition, 1988, page 148, has `x-cu', where
> Jones, Sato, Wada, and Wiens in their American Math Monthly article,
> 1976, page 449, have `x+cu'. Does anyone know whether Ribenboim's
> `x-cu' is a correction or a misprint?
Sharp eyes!
Ribenboim 2nd ed has it corrected ...+1-(x+cu)^2]^2...
Professor Caldwell of UTM has kindly volunteered to be an official host
of the errata to the The New Book of Prime Number Records 2nd edition:
http://primepages.org/notes/errata/
Phil
Raymond Manzoni
>
> Does anyone have values for the 26 variables that actually produce a
> positive prime from this formula? (like for k = 0) I tried to
> produce one for a few hours with no success. Is it that the variables
> have to be unreasonably huge to do it?
>
> Mike Bales
The point is that the nonnegative variables have to verify at once the
14 conditions (that is All the squares appearing in the polynomial must
be 0, otherwise the result couldn't be positive) :
(1) q=w*z+h+j
(2) z=(g*k+g+k)*(h+j)+h
(3) f^2=16*(k+1)*k^3*(n+1)^2+1
(4) e=p+q+z+2*n
(5) o^2=e^3*(e+2)*(a+1)^2+1
(6) x^2=(a^2-1)*y^2+1
(7) u^2=16*(a^2-1)*r^2*y^4+1
(8) (x+c*u)^2=(a+u^2*(u^2-a))^2-1)*(n+4*d*y)^2+1
(9) m^2=(a^2-1)*l^2+1
(10) l=k+i*(a-1)
(11) y=n+l+v
(12) m=p+l*(a-n-1)+b*(2*a*(n+1)-(n+1)^2-1)
(13) x=q+y*(a-p-1)+s*(2*a*(p+1)-(p+1)^2-1)
(14) p*m=z+p*l*(a-p)+t*(2*a*p-p^2-1)
Let's try to find the variables for k=0 (delivering the prime 2)
If ff(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v,
w, x, y, z) =
(k+2)*(1-((w*z+h+j-q)^2+((g*k+g+k)*(h+j)+h-z)^2+(16*(k+1)*k^3*(n+1)^2+1-f^2)^2+(2*n+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+(16*r^2*y^4*(a^2-1)+1-u^2)^2+(((a+u^2*(u^2-a))^2-1)*(n+4*d*y)^2+1-(x+c*u)^2)^2+((a^2-1)*l^2+1-m^2)^2+((a-1)*i+k-l)^2+(n+l+v-y)^2+(p+l*(a-n-1)+b*(2*a*n+2*a-n^2-2*n-2)-m)^2+(q+y*(a-p-1)+s*(2*a*p+2*a-p^2-2*p-2)-x)^2+(z+p*l*(a-p)+t*(2*a*p-p^2-1)-p*m)^2))
then I got :
ff(20,0,0,0,3,1,0,1,0,0,0,0,1,0,244,1,1,0,0,0,1,0,0,1,0,1)=2
(that is a=20,b=0,...)
Other solutions exist, for example :
ff(20,0,0,0,3,1,1,0,0,1,0,0,1,0,244,1,1,0,0,0,1,0,0,1,0,1)=2
(what follows is far from a nice proof but merely the draft of my
search of at least one solution. When more than one choice was available
I choose the smallest one)
k=0
f=1
p=1
q=2^n
z=1
(1) 2^n=w+h+j
(2) 1=g*(h+j)+h
(4) e=2+2^n+2*n
(5) o^2=e^3*(e+2)*(a+1)^2+1
(6) x^2=(a^2-1)*y^2+1
(7) u^2=16*(a^2-1)*r^2*y^4+1
(8) (x+c*u)^2=(a+u^2*(u^2-a))^2-1)*(n+4*d*y)^2+1
(9) m^2=(a^2-1)*l^2+1
(10) l=i*(a-1)
(11) y=n+l+v
(12) m=1+l*(a-n-1)+b*(2*a*(n+1)-(n+1)^2-1)
(13) x=2^n+y*(a-2)+s*(4*a-5)
(14) m=1+(l+2*t)*(a-1)
a=20
n=0
q=1
e=3
o=244
y in {0,1,40,1599,..}
l in {0,1,40,1599,..}
w=0
h=1
j=0
g=0
b=t
(1) 1=w+h+j
(2) 1=g*(h+j)+h
w=0 (if w=1 h+j=0 impossible)
(1) 1=h+j
(2) 1=g+h
h=1 or h=0
j=0 j=1
g=0 g=1
b=t = 0
(6) x^2=399*y^2+1
(7) u^2=16*399*r^2*y^4+1
(8) (x+c*u)^2=(20+u^2*(u^2-20))^2-1)*(4*d*y)^2+1
(9) m^2=399*l^2+1
(10) l=i*19
(11) y=l+v
(13) x=1+y*18+s*75
(14) m=1+(l+2*t)*19
l=0
m=1
t=0
i=0
y=v (=0 latter)
(6) x^2=399*y^2+1
(7) u^2=16*399*r^2*y^4+1
(8) (x+c*u)^2=(20+u^2*(u^2-20))^2-1)*(4*d*y)^2+1
(13) x=1+y*18+s*75
I'll try y=0
x=1
u=1
s=0
(8) (1+c*u)^2=1
c=0
The unused variables are considered zero.
Hoping it's not to wrong,
Raymond
Raymond Manzoni
>
> 14 conditions (that is All the squares appearing in the polynomial must
> be 0, otherwise the result couldn't be positive) :
Well at least the most exterior squares...
Raymond
hack
I wouldn't be surprised if some of the variables turned out to be quite
large; after all, they're sort of Goedel numbers for formulae expressing
various propositions P(x), Q(x) whose conjunction expresses the fact that
x is prime. The complete polynomial is basically x(1-P#2(x)-Q#2(x)...)
which, for x>0, has the value p for x=p iff P(p)=0 and Q(p)=0 etc., and
is negative otherwise. (Offhand I don't remember how to handle the case
of nonpositive x.) The trick was to express each of those propositions
as a polynomial equation involving x and auxiliary variables.
(Don't yell at me if the particular 26-variable polynomial is not precisely
of this form -- I described the general mechanism for constructing things
like this.)
Michel.
gs...@sky.com
Raymond, I tried your values in an Excel file but still kept getting a negative answer, for example the 16(k+1)^3(k+2)(n+1)^2+1-f^2 bit makes 32?
Spac...@hotmail.com
Another theorem proved in this (1976) paper was that the number of variables in the prime representing polynomial can be reduced to 12 (that the set of primes can be defined in 11 unknowns). This result has been superceded. Primes can be diophantine defined in 9 unknowns. Hence there exists a exists a prime representing polynomial in 10 variables. (See paper 15.)