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I see this page (https://en.wikipedia.org/wiki/Category:Classes_of_prime_numbers) about different forms of prime numbers. I see also, that it was conjectured that all these forms contain infinitely many primes.

My question is about the existence of a mathematical form (expression) but with a finite number of primes.

Safwane
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    Does $2n$ qualify? – lhf Jul 24 '19 at 16:36
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    The class of all primes less than 2000 is a finite class of primes. – fleablood Jul 24 '19 at 16:37
  • @lhf: Yes, this is an example. – Safwane Jul 24 '19 at 16:38
  • That isn't a list of different forms of prime numbers but different sets of prime numbers with certain reasons to group them together as a class. There are infinitely many ways to take sets of prime numbers many very boring. Any finite set of prime numbers is a finite set of prime numbers so.....Even prime numbers. That's finite. Prime numbers that are equal to either 3, 7 or 41. That's finite. Etc. – fleablood Jul 24 '19 at 16:40
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    Perhaps a better question is this: Is there an irreducible polynomial with integer coefficients that represents only finitely many primes. See https://math.stackexchange.com/questions/169066/polynomials-representing-primes – lhf Jul 24 '19 at 16:41
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    Okay.... it seems you are looking at a wikipedia page entirely different to the one you posted..... – fleablood Jul 24 '19 at 16:41
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    A more interesting example is $x^2+x+2$ which is always even although the polynomial is irreducible. As formulated, the question is too broad. Precise "form" – Peter Jul 24 '19 at 16:43

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There are many such examples...

Say for example $p\cdot n$, where $p$ is any prime number. To be more precise, the functions $2n,3n,5n,7n,11n,\ldots$ are the ones, which are prime only when $n=1$.

Anand
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Consider primes of the following forms:
(A): $p=$ any reducible polynomial, $p=f(x)\cdot g(x)$ each one having integer coefficients.
If it represents infinitely many primes, then one of them (say $f(x)$) must take the values $1$ or $-1$ infinitely often which is impossible.
(B): $p\pmod{p_1}, p\pmod{p_2}, \ldots p\pmod{p_k}$ are all powers of $2$ where $p_1, p_2, ..., p_k$ are the primes less than $p$.

(This is not so easy to see that there are a finite number of such primes $p$, but here is a quick proof).
If this happens, then $p_1\cdot p_2\cdots p_k$ divides $(p-1)(p-2^1)(p-2^2)\cdots (p-2^x)<p^{x+1}$ where $2^x<p<2^x+1$ which means that $x<\frac{\ln p}{\ln 2}$.
This shows that $p_1\cdot p_2\cdots p_k\le p^{x+1}<p^{\frac{\ln p}{\ln 2}+1}$. Taking logarithms in both sides we get that $\ln(p_1\cdot p_2\cdots p_k)< (\frac{\ln p}{\ln 2}+1)\cdot \ln p \Rightarrow \theta(p)=\mathcal{O}(\ln^2p)$.
Here $\theta(p)$ denotes the usual Chebyshev function which is known to be $\mathcal{O}(p)$ which contradicts the previous argument.
However, there are some primes satisfying the claim, such as $3, 5, 7, 11$.

Konstantinos Gaitanas
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Any integer polynomial, with an even constant term, and an even number of odd coefficients in both paired up and unpaired sections ( parity arguments show this) ex:$$3x^{10}y^5+5x^9y^4+7x^8y^3+11x^7y^2+20x^6y+23x^5+12x^4y^7+8x^3+ 200x^2+21x+2$$ Will never regardless of $x$ or $y$'s values, be anything but even. Because, odd coefficient terms can pair up if both variables are odd ( or just $x$) , leaving none unpaired. If $x$ is even then all the terms with it are even. And if $y$ is even then all odd coefficient terms without it, can pair up if $x$ is odd, or be on their own if both are even.