For different integers there exists an integer so that exactly one of the sums is prime. Let $i,j$ be distinct natural numbers. In a comuter science class the lecturer said

Since the prime numbers are not regularly distributed, there is some natural number $k$ such that $i+k$ is prime and $j+k$ is not.

I found this reasoning a bit unsatisfying, so I tried to show the claim, however it turned out to be not that trivial. If lets say $i$ has a prime factor $p$ that doesn't divide $j$ then by Dirichlet there is a prime $q\equiv j\pmod p$, $q>j$. Then choose $k=q-j$, so $j+k$ is prime but $i+k$ has the prime factor $p$, hence is not prime. But this reasoning doesn't work if $i,j$ have the same prime factors. I also feel like this can be proven more elementary but I don't see how.
Edit: Thanks everyone for the help, I learned something from every answer. Unfortunately I can only accept one of them, so I hope I didn't offend anyone for not accepting their answer.

Let $i$ and $j$ be distinct positive integers and assume that there is some positive integer $k$ such that $j+k$ is prime whenever $i+k$ is prime.
Assume $i<j$
Then, for every prime number $p$ larger than $i$ which can be written as $i+(p-i)$ , the number $$j+(p-i)=p+(j-i)$$ is prime as well. Since $j-i$ is fixed, we can establish an upper bound for all prime gaps which contradicts the fact the prime gaps can become arbitarily large.
The case $i>j$ gives the same result : From some point on, the prime gaps are bounded by $i-j$ , since for every sufficient large prime $p$ , $p-(i-j)$ is prime as well.
