Diophantine representation of the set of prime numbers of the form $n²+1$ 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.

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 $N^2+1$.
