The integer $n$ is not zero if and only if there is some prime $p>n$ such that $p-n$ is composite I recently solved a problem (it was posed in a spanish-talking forum) in which I used the following lemma.

The integer $n$ is not zero if and only if there is a prime $p>n$ such that $p-n$ is composite.

This is my proof (I only write the "non-trivial" implication):

Let $n$ be a non-zero integer and let $p>|n|$ be a prime. Then there are integers $q$ and $r$ such that $$n=pq+r$$ and $0<r<p.$ It follows that if $p'>n+p$ is a prime of the form $p'=pk+r,$ $k\in\mathbb Z$ (its existence follows from Dirichlet's Theorem) then $$p'-n=pk+r-pq-r=p(k-q)$$ and since $k-q>1,$ $p'-n$ is a composite number (note that in fact there are infinitely many such primes). 

My question is, can that result be proved in a more elementary way (without appealing to Dirichlet's theorem)?

Let $n\neq 0$ and $q>n$ be prime. Consider $x=q-n$, now if $n$ be an odd integer, then $q-n$ is even and so $x$ is composite number. Now let $n$ an even integer. then $n=q-x$ now if $\gcd(x,q)= 1, q$, since $x\neq q$ so if $\gcd(x,q)=q$ then $q|x$ i.e., $x$ is composite number, so let $\gcd(x,q)=1$, then there is integers $r ,s$ such that $rx+sq=1$ and so $(ns-1)q=(-nr-1)x$ now if is a prime then $ns-1=x$ and $q=-nr-1$, therefore $$n=-nr-1-ns+1=n(-r-s);$$
$$-r-s=1\rightarrow r=1-s \rightarrow s(q-x)=1 \rightarrow s=1, r=0$$
where this implies that $q=1$, but $q$ is prime, thus $x$ can not be a prime and proof is done.
