## Today I learned

Posted by – June 7, 2010

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-f2]2
[e3(e+2)(a+1)2+1-o2]2 – [(a2-1)y2+1-x2]2 – [16r2y4(a2-1)+1-u2]2
[((a+u2(u2-a))2 -1)(n+4dy)2 + 1 – (x+cu)2]2 – [n+l+v-y]2 – [(a2-1)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]

##### 4 Comments on Today I learned
1. Neil Hardwick says:

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

2. sam says:

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!

3. Vadim Kulikov says:

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

4. sam says:

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.