Does there exist an f(x) minimum first-degree polynomial with integer coefficients, such that for any x integer number, f(x) is prime?
Any help would be appreciated!
No!
Proof by reductio ad impossibile.
Suppose that there is exist such polynomial $\displaystyle f(n)$.
$\displaystyle f(n)=a_kn^k+...+a_1n+a_0$
Let now $\displaystyle n=n_0$ such that $\displaystyle f(n_0)=p$, when $\displaystyle p$ is prime.
Let now be $\displaystyle t\in\mathbb{N}$, and let us look at:
$\displaystyle f(n_0+tp)=a_k(n_0+tp)^k+...+a_1(n_0+tp)+a_0$
$\displaystyle =a_kn_0^k+...+a_1n_0+a_0+pQ(t)$
$\displaystyle =f(n_0)+pQ(t)$
$\displaystyle =p+pQ(t)=p(1+Q(t))$
$\displaystyle Q(t)$ is polynomial with integer coefficients.
So, $\displaystyle p\mid{f(n_0+tp)}$, but $\displaystyle f(n)$ is generates primes only, therefor $\displaystyle f(n_0+tp)=p$ for all integer $\displaystyle t$.
Now, from the fact that polynomial can't have the same value more than $\displaystyle k$ times, we get the contradiction!
Here some interesting polynomials:
$\displaystyle f(n)=110,437+13,860n$ when $\displaystyle n=0,1,...,10$ then $\displaystyle f(n)$ is prime...
Also(famous one):
$\displaystyle f(n)=n^2+n+41$
There are many such functions. For example, Riemann gave an explicit formula for the number of primes less than $\displaystyle x$. A prime-yielding formula can be easily constructed from this.
There is also a positive constant $\displaystyle A$, known as Mill's constant, such that $\displaystyle \lfloor A^{3^n}\rfloor$ is prime for every $\displaystyle n$. It's about $\displaystyle 1.306...$ if the Riemann hypothesis is correct. It yields the sequence of primes $\displaystyle 2, 11, 1361, 2521008887, ...$.
Also, the simple recurrence relation $\displaystyle a_1=7, a_{n+1}=a_n+\mbox{gcd}(a_n,n)$ generates only primes.
There is no reason why any given kind of "formula for primes" could not exist. It is a popular belief that there is no such thing; I've even heard non-mathematicians claim it, as if it were some kind of established result. I don't believe the OP was asking a silly question at all, because there do exist many such formulae.
Here's another proof that no non-constant polynomial with integer coefficients yields only primes. Let $\displaystyle c=f(0)$ be the constant term of this polynomial. It's quite easy to see that $\displaystyle f(mc)$ is divisible by $\displaystyle c$ for every integer $\displaystyle m$.