For every pair of distinct primes $p_1$ and $p_2$, there are $z$ such that $p_1 + z$ is prime but $p_2 + z$ is not

Question

Just a curiosity that occurred randomly. Is there a proof of the following:

For every two distinct primes $p_1$ and $p_2$, there exist an infinity of positive integers $z$ such that $p_1 + z$ is prime, but $p_2 + z$ is not prime.

It just "feels" true, but I don't know enough number theory to get an idea on how to prove this. I can't even guarantee that one such $z$ always exists.

Answer 1

Apply Dirichlet's Theorem on prime numbers in arithmetic progression. For $z$ try all possible multiples of $p_1$. The $z+p_1$ will always be composite; but $p_2+xp_1$, for various positive integers $x$ represents an arithmetic progression of common difference $p_1$, with $p_2$ as first term as $\gcd(p_1,p_2)=1$ we are done.

Answer 2

Let $r=p_2-p_1$. There are infinitely many primes $p$ such that $p+1,p+2,\ldots,p+r$ are composite: take for example the greatest prime lesser than $s!+2$, for each $s>r$.

Answer 3

Similar to ajotatxe's answer but a little more elfficient construction. Let $r = p_2-p_1$. Suppose initially that $r>0$. Let $p$ be any prime $> p_1$. Clearly $p+kr$ is not always prime for every choice of $k$ (for instance take $k=p$). Suppose $p+k_0r$ is the smallest composite value, then take $z = p + (k_0-1)r - p_1 = p + k_0r - p_2$.

For the case $r<0$, the construction is similar but we would pick a small prime $q$ that doesn't divide $r$, start at a prime $p > p_2 + qr$, and then count backwards by steps of $r$.