We'll use Dirichlet theorem: there's infinitely many primes of the form $a + nd$ if $gcd(a,d)=1$. Firstly, we may assume $x$ is prime : we start asking if $x+1,x+2,\dots$ is prime until we reach a yes answer, and set that $x+c$ as our new $x$. Now, for each prime $p$ we find prime numbers $q_1,\dots, q_{p-1}$ bigger than $2p$ for which $q_i \equiv i \mod p $. Imagine we ask if $x+q_i - p$ is prime. If $x=p$ we will get all yes answers. If $x\neq p$ then $x\equiv -i \mod p$ for some $i$ in $1,\dots p-1$ (since $x$ is prime), and so $p \mid x+q_i-p > p$ is not prime, and one of the answers will be no. We therefore can run this for all prime values $p$ until reaching all yes answers.