Guess what? For variables (a to z) ranging over the nonnegative integers, the set of positive values of this polynomial is the set of prime numbers:

(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} – [(a^{2}-1)y^{2}+1-x^{2}]^{2} – [16r^{2}y^{4}(a^{2}-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} – [(a^{2}-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}]

Funny thing – Bess and I were just talking about this very same thing here in the Green Parrot in Key West!

Neil Hardwick

Heh. The Green Parrot even sounds like my kind of place, judging from what Bess told me about it. Hope you’re having a great time!

sam

And the eighth power is the largest I can see? Wow, can you prove it? A lot of variables though..

Vadim Kulikov

I can’t, but it’s based on

1) the representability of diophantine systems as polynomials

2) the formulation of primes as such a system (eg. here)

If you inspect the polynomial closely, you notice that it has the form

(k+2) * (1 – P(a,b,c…)^2 – P'(a,b,c…)^2 …), so it will be positive only when all the polynomials P are zero. They essentially select the prime values of k+2. More on Wikipedia.

sam