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.
 A: Suppose $i > j$. Further suppose that the statement is not true. Then there are two cases:
Case 1: $i + k$ is not prime for all $k\in \mathbb N$. This contradicts the infinitude of primes.
Case 2: $i+k$ is prime for some $k\in\mathbb N$ if and only if $j+k$ is also a prime. In that case:
Since $j + (k + i - j)$ is prime, so is $i + (k+i-j) = 2i -j + k$.
Since $j + (k + 2i - 2j)$ is prime, so is $i + (k + 2i - 2j) = 3i - 2j + k$.
Since $j + (k + 3i - 3j)$ is prime, so is $i + (k + 3i - 3j) = 4i - 3j + k$, and so on...
We have now created an infinite arithmetic progression of primes: $i + k + n(i-j)$ is prime for all $n \in \mathbb N$. To see why this is a contradiction, consider the remainders of this progression when dividing by $i-j+1$.
A: Suppose $i<j$. Add a number to each to make $j+k=p$, a prime.
Now $i+k$ is coprime to $p$ and so by Dirichlet there is a number $l$ such that $i+k+lp$ is prime, whereas $j+k+lp$ is a proper multiple of $p$.
A: An elementary proof
Suppose $i<j$ and let $p$ be any prime greater than $i$. Let $a=j-i$.
Consider the sequence $p,p+a,p+2a,...,p+pa$. The initial number is prime and the final number is composite and so there is a successive pair of these numbers $u<v$ where precisely one is prime.
Then $k=u-i$ solves the problem.
A: 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.
