A polynomial formula for the primes (with 26 variables) was presented by Jones, J., Sato, D., Wada, H. and Wiens, D. (1976). Diophantine representation of the set of prime numbers. *American Mathematical Monthly*, *83*, 449-464.


The set of prime numbers is identical with the set of positive values taken on by the polynomial


$(k+2)(1-(wz+h+j-q)^2-((gk+2g+k+1)\cdot(h+j)+h-z)^2-(2n+p+q+z-e)^2-(16(k+1)^3\cdot(k+2)\cdot(n+1)^2+1-f^2)^2-(e^3\cdot(e+2)(a+1)^2+1-o^2)^2-((a^2-1)y^2+1-x^2)^2-(16r^2y^4(a^2-1)+1-u^2)^2-(((a+u^2(u^2-a))^2-1)\cdot(n+4dy)^2+1-(x+cu)^2)^2-(n+l+v-y)^2-((a^2-1)l^2+1-m^2)^2-(ai+k+1-l-i)^2-(p+l(a-n-1)+b(2an+2a-n^2-2n-2)-m)^2-(q+y(a-p-1)+s(2ap+2a-p^2-2p-2)-x)^2-(z+pl(a-p)+t(2ap-p^2-1)-pm)^2)$


as the variables range over the nonnegative integers.


I am asking if there exist a similar result for the primes of the form $n²+1$, i.e., this set of prime numbers is identical with the set of positive values taken on by certain polynomial.



Diophantine representation of the set of prime numbers of the form $n²+1$
Call your polynomial $P$. I propose the following polynomial:
$$
P' = (\xi^2+1)(1 - (\xi^2+1-P)^2)
$$
**Proof** (that the positive values of $P'$ are exactly the primes of the form $N^2+1$): 


Let $P\_0$ be one of the values of $P$, and let $\xi\_0$ be any integer. 


Case (i). Suppose $P\_0 = \xi\_0^2+1$. Then the value of the above polynomial (with the appropriate substitutions made) is $P\_0 = \xi\_0^2+1$, since the second factor evaluates to unity.


Case (ii). Suppose $P\_0 \neq \xi\_0^2+1$. Now the second factor evaluates to some non-positive number, and hence the value of the polynomial itself is non-positive.


Now the first case gives us *all* primes of the form $N^2+1$ as values of $P'$, and no other values (since $P\_0=\xi\_0^2+1$ is positive), whereas the second case gives us zero or negative numbers exclusively. So the positive values of $P'$ are exactly the primes of the form $N^2+1$.


