## Searching for solutions of Jones polynomial

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-f2)2 - (e3(e+2)(a+1)2+1-o2)2 - (y2(aa-1)+1-x2)2 - (16r2y4(aa-1)+1-u2)2 - (((a+u2(u2-a))2-1)(n+4dy)2+1-(x+cu)2)2 - (n+l+v-y)2 - ((aa-1)l2+1-m2)2 - (ai+k+1-l-i)2 - (p+l(a-n-1)+b(2an+2a-n2-2n-2)-m)2 - (q+y(a-p-1)+s(2ap+2a-p2-2p-2)-x)2 - (z+pl(a-p)+t(2ap-p2-1)-pm)2

Let's try to find {a,b,...,z} values for prime 2.
P can be written (k+2)(1-A˛-B˛-C˛-...-N˛). P is prime if (1-A˛-B˛-C˛-...-N˛)=1, <=> A˛=B˛=C˛=...N˛=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.

 a,b,c...,z variables (check box to freeze value) {a,b,...,z}: A,B,...N equations
random maximum (as a power of 10, ex : enter 3 for random numbers from 1 to 103=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: