Partitions of primes vs. irreducible polynomials

Question

I need help to look for counter-examples of a conjecture (reformulation nr 2):

$f\in\mathbb Z[X_1,\dots,X_n],\;n>1,$ not independent of any variable and not such that $f(X_1,\dots,X_n)(X_1+\cdots +X_n)\equiv 0\pmod 2$ and not possible to factorize with integer valued polynomials* in $\mathbb Q[X_1,\dots,X_n]$ is irreducible iff for all primes big enough $p$ there exists a partition $p=\sum N_k$, $N_k>0$, such that $f(N_1,\dots,N_n)$ is prime.

I have tested some irreducible polynomials but have found no exception. The background is mostly the question

Any odd number is of form $a+b$ where $a^2+b^2$ is prime

where the polynomial $X^2+Y^2$ was tested up to $10^8$.

I have tested
$x^2+y^2+z^2$
$x^3+y^3+z^3$
$xy+yz+zx$
$x^3+y^3+z^3-3xyz$
$x^2+yz+z$
$x+y+z^2$
$x^5+y^5+z^5$
$x^5+y^5-z^5$
$x^3-y^2+xz$
$x^2+y^2+z^2+xy+yz$
$x^3+y^3+z^3+xyz$
$x^2+x+yz$

No meaning testing reducible polynomials, since the reverse part is obvious.

I'm asking ChatGPT for irreducible polynomials, but if someone know a more secure source I would be glad to know it.

(*) A polynomial $f\in\mathbb Q[X_1,\dots,X_n]$ that is a restriction $\mathbb Z^n\to \mathbb Z$.

Answer 1

If a polynomial $f\in \mathbb Z[X_1,\dots,X_n]$ multiplied with $X_1+\cdots+X_n$ is equivalent with $0\pmod 2$ then all values of $f$ for a prime $X_1+\cdots+X_n$ will be even (and likely not prime). To see that think of multiplication as $and$ and addition as $exclusive\; or$.

Example, algebra in a Boolean ring:
$(xyz+xy+yz+xz)(x+y+z)=\\ x^2yz+x^2y+xyz+x^2z+xy^2z+xy^2+y^2z+xyz+xyz^2+xyz+yz^2+xz^2=\\ 3xyz+x^2yz+xy^2z+xyz^2+x^2y+x^2z+xy^2+xz^2+y^2z+yz^2=6xyz+2xy+2yz+2xz=0$

One could replace the condition of non constant terms with a condition that $f(X_1,\dots,X_2)(X_1+\cdots+X_n)\pmod 2$ not should be equivalent to $0$, but there could be other obstructions for other modular calculations.

The last sentence is questionable since $\mathbb Z_p[X_1,\dots,X_n]$ is an integral domain for primes $p>2$.