For different integers there exists an integer so that exactly one of the sums is prime.

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 $i-j+1$.

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 $p$.

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.