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January 29

Example of Jones formula

If we let the 26 variables be a=1, b=2, c=3, …, z=26, then what are the solutions of the set of 14 Diophantine equations? 118.170.15.127 (talk) 11:16, 29 January 2025 (UTC)[reply]

The first equation of the set of 14 equations, is:

wz+h+j-q=0.

Using the assignment of values

w=23,z=26,h=8,j=10,q=17,

this becomes

599=0.

Seeing as this ain't so, the system has no solutions.

If the set had a solution under this specific assigment, it would be this:

a=1,b=2,c=3,...,z=26.

The question is a bit like, "letting

x=1,

solve the equation

2x=4.

" --Lambiam 12:28, 29 January 2025 (UTC)[reply]

You can't just take a random set of values and solve for x - there is no x in the usual sense of single variable polynomial in the formula - one must think of all a to z as 26 different x's. What the Jones formula does is provide one with a way of proving a number is prime by supplying 26 numbers and showing the result of that formula is the prime number. Which is quite amazing - one just needs to do a small constant number of operations - addition subtraction multiplication and comparisons with zero. However the numbers can be of the order of the prime to the power of itself - so definitely not practical to generate never mind use! NadVolum (talk) 13:09, 29 January 2025 (UTC)[reply]

There is a prime generating polynomial of Jones formula:

{\begin{aligned}&(k+2)(1-{}\\[6pt]&[wz+h+j-q]^{2}-{}\\[6pt]&[(gk+2g+k+1)(h+j)+h-z]^{2}-{}\\[6pt]&[16(k+1)^{3}(k+2)(n+1)^{2}+1-f^{2}]^{2}-{}\\[6pt]&[2n+p+q+z-e]^{2}-{}\\[6pt]&[e^{3}(e+2)(a+1)^{2}+1-o^{2}]^{2}-{}\\[6pt]&[(a^{2}-1)y^{2}+1-x^{2}]^{2}-{}\\[6pt]&[16r^{2}y^{4}(a^{2}-1)+1-u^{2}]^{2}-{}\\[6pt]&[n+\ell +v-y]^{2}-{}\\[6pt]&[(a^{2}-1)\ell ^{2}+1-m^{2}]^{2}-{}\\[6pt]&[ai+k+1-\ell -i]^{2}-{}\\[6pt]&[((a+u^{2}(u^{2}-a))^{2}-1)(n+4dy)^{2}+1-(x+cu)^{2}]^{2}-{}\\[6pt]&[p+\ell (a-n-1)+b(2an+2a-n^{2}-2n-2)-m]^{2}-{}\\[6pt]&[q+y(a-p-1)+s(2ap+2a-p^{2}-2p-2)-x]^{2}-{}\\[6pt]&[z+p\ell (a-p)+t(2ap-p^{2}-1)-pm]^{2})\\[6pt]\end{aligned}}

Does this polynomial generate a prime number if a=1, b=2, c=3, …, z=26? --114.38.87.55 (talk) 07:55, 30 January 2025 (UTC)[reply]

This polynomial is found in the article. The answer to the question is no. It only produces a nonnegative value if all 14 Diophantine equations are satisfied. As you can read above, with the given value assignment, it fails already on the first equation. --Lambiam 11:14, 30 January 2025 (UTC)[reply]

So what number does this polynomial generate if a=1, b=2, c=3, …, z=26? 111.252.80.160 (talk) 11:36, 30 January 2025 (UTC)[reply]

It'll produce a negative number even though k+2 is 13. All those squares in the big second term need to be zero otherwise it produces zero or a negative number instead of 13 x 1. NadVolum (talk) 12:13, 30 January 2025 (UTC)[reply]