According to Jones paper, the set of prime numbers is identical with te set of positive values taken on by the following polynomial, this page helps finding {a,b,...,z} values that match it.

(1) P = (k+2)(1 - (wz+h+j-q)^{2}
- ((gk+2g+k+1)(h+j)+h-z)^{2}
- (2n+p+q+z-e)^{2}
- (16(k+1)^{3}(k+2)(n+1)^{2}+1-f^{2})^{2}
- (e^{3}(e+2)(a+1)^{2}+1-o^{2})^{2}
- (y^{2}(aa-1)+1-x^{2})^{2}
- (16r^{2}y^{4}(aa-1)+1-u^{2})^{2}
- (((a+u^{2}(u^{2}-a))^{2}-1)(n+4dy)^{2}+1-(x+cu)^{2})^{2}
- (n+l+v-y)^{2}
- ((aa-1)l^{2}+1-m^{2})^{2}
- (ai+k+1-l-i)^{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}

P can be written (k+2)(1-A²-B²-C²-...-N²). P is prime if (1-A²-B²-C²-...-N²)=1, <=> A²=B²=C²=...N²=0. This page let us try to solve each of these equations for P=2.

P=2 => (k+2)=2, A=0, B=0,...N=0 => k=0, A=0, B=0,...N=0. First 20 variables were found and described in Gupta's thesis. 6 are missing.

^{3}=1000):nb Calc:0 |

This part is for unit tests on computations

Search for a r value that would give a u/uu=16rryyyy(aa-1)+1

r | |

16rryyyy(aa-1)+1 | |

sqrt(uu) |

x: | |

formula: | |

result: |