Let $f(x)$ be an irreducible integer polynomial of degree $k$. Let $x_1,x_2,...,x_j$ be some zeros of $f(x)=0$ where $j<k$.
How do I find identities of type $P(x_1,x_2,...,x_j) = 0$ where $P$ is an integer polynomial that has at most degree $m<k$ in every variable $x_1,x_2,...x_j$ ?
Im not an expert in Galois theory , group theory , ring theory or Hamil basis.
How do I find such identities without being an expert in Galois theory and without finding a closed form for one of the zero's of $f(z)=0$ ?
Example : I got some degree 11 integer polynomial and I want to find an identity like $17 x_1 x_2 + 5 x_1 x_3 + 4 x_2 x_3 +5 = 0$
Im not sure if this is the most efficient method but by lack of another answer I give a method that will work to detect if the considered identity could be correct.
I would love to see a method based on calculus if that is possible !! Anyway ...
Let the given polynomial $f(x)$ be of the form $f(x) = a_k x^k + a_{k-1} x^{k-1} + ... + k_0$.
Let $A$ be $max_i(|a_i|)$ ; thus the largest integer coëfficiënt in $f(x)$ in terms of absolute value.
Let $B = 2A(k+1) + 1$.
Let $P_t$ be a prime larger then $B$ such that there is a solution for $x$ in $f(x) = 0 \mod P_t$.
Then if the identity $P(x_1,x_2,...,x_n) = 0$ is true then it is also true mod $P_t$.
SO we have a neccessary condition.
Notice the number of potential identities is finite because $x_i^{k+1}$ can be reduced and also large integers can be reduced by using Vieta's formula's and the discriminant.
Therefore by checking all possibilities with the above modular method we can eliminate many false identities.
Im unaware of any simpler method or any calculus method.
Well at least for general case situations.
With special thanks to my mentor.
EDIT
I replaced the claimed "iff" with the weaker "if". In other words I realized its just a neccessary condition but not sufficient. I assume that mistake created the downvotes.
The full answer , thus the "IFF test" is probably testing the identities on multiple primes $P_t$. But it's not completely clear to me yet.