$(z-k)$ is composite then $(z-1)+(k-1)$ is also composite(A proof for composite number).

Question

Given

1. $z(z-1)$ is divisible by all prime $< n$ where $ n>\sqrt z$
2. $(z+k)$ is prime.

Prove or disprove if $(z-k)$ is composite then $(z-1)+(k-1)$ is also composite.

Proof

$z-k=(z-1)-(k-1)=E$, where $E$ is a composite number, consider $e$, where $e | E$. If $\gcd((z-1),(k-1))=g>1$ then $(z-1)+(k-1)$ is composite.

If $\gcd((z-1),(k-1))=1$ then there is no such $e \mid E$, so that $e \mid E$ and $e \mid (z-1),(k-1)$ at the same time.
$e \not\mid z, k$ since $(z+k)$ is prime, which implies $(z,k)=1$. So, $e \not\mid z, k, (z-1),(k-1)$.
But $E$ must have a prime $< E^{1/2} =(z-k)^{1/2} <z^{1/2}<n .$ So, $((z-1),(k-1))>1 $since $((z-1),(k-1))=1$ does not allow $E$ to have any prime of $z,(z-1)$. So, if $(z-k)$ is composite then $(z-1)+(k-1)$ is also composite.

** this mundane problem might have a relation to brocard's problem!! any kind of comment would help.

Answer 1

The proof you give is a bit unstructured and difficult to follow, and as a result not very convincing. However, the underlying idea seems to be correct. Here is almost the same proof, but rendered differently.

Assume $z$ and $k$ with $k < z$ are positive integers such that $z + k$ is prime, $z - k$ is composite, and any prime number smaller or equal to $\sqrt z$ divides $z(z-1)$. We want to prove that $(z - 1) + (k - 1)$ is composite.

Let $p$ be the smallest prime dividing $z - k$. Since $z - k$ is composite, it follows that $p \leq \sqrt{z - k} < \sqrt z$. By the assumption on $z$, it follows that $p$ divides $z(z -1)$. Since $p$ divides $z-k$ and does not divide $z+k$, $p$ does not divide $z$, and so we conclude that $p$ divides $z - 1$.

But then $p$ divides both $z-k$ and $z - 1$, so it also divides $k- 1$, and therefore also $(z -1) + (k - 1)$. In particular this last number is composite, and this is what we wanted to prove.

As a final remark, I would like to add that the assumption that any prime smaller than $\sqrt z$ divides $z(z - 1)$ forces $z$ to be smaller than $17^2 = 289$, since otherwise the product of all primes smaller than $\sqrt z$ is already (much) larger than $z(z-1)$. So one could also check all pairs $(z, k)$ satisfying your conditions with a computer.