Let $k>3$ be an integer. Show that it's not possible fir k prime numbers each greater than $k$ to be in arithmetic progression with a common difference less than or equal to $k+1$.
The process I'm trying is by assuming $k$ is odd and not divisible by $3$ in one case and other such assumptions but it is proving to be a very lengthy process. I am looking for a simpler approach to solve this.
Let $d$ be the common difference. Obviously $d\neq 1$, and for now assume $d\neq 2$.
Look at the list mod $d-1$. By assumption the list looks like $a,a+1,a+2,..., a+k+1$ where $a$ is congruent to the first prime in the list, since their successive differences are 1 mod $d-1$. But because $k+1>d-1$ (since $d\leq k+1$) we are guaranteed that $0$ appears in that list, simply that means because it has to cycle through all the integers mod $d-1$. So one of the primes is congruent to zero mod $d-1$ and hence must equal $d-1$ because it’s a prime. But by assumption, all the primes are greater than $k$ but $d-1\leq k$. That’s a contradiction, so there are no such primes.
Now for $d=2$, notice that the above argument still works with $d+1=3$ instead of $d-1$ because we assumed $k>3$.