Originally Posted by Jones et al
In this paper it is proved that the set of prime numbers is exactly the set of positive values of a polynomial of the 25th degree, in 26 variables as the variables range over the nonnegative integers:
(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}
Each positive value of the above polynomial is prime. Also each prime is a value of the above polynomial, for some nonnegative integers a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y, z. Thus the set of prime numbers is identical with the set of positive values of the polynomial.
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I may be missing something obvious here but surely this function is the result of multiplying two numbers (k+2) and {} together and hence can't be prime unless (k+2) equals 1 (or {} equals 1.) If we set k+2 = 1 then k=-1 and a cursory examination of the second (larger) section of the formula shows that in most cases 'k' appears close to a '+1)
Although the formula is quoted by de sautoy (music of the primes) this fact (+ the incredible coincidence that the number of variables 'just happens' to be the same as the number of characters in the alpahabet) suggests someone is playing a practical joke.
Can someone enlighten me?
