Show that if $n$ and $k$ are positive integers, with $n>1$ and all $n$ positive integers $a, a+k,..., a+(n-1)k$ are odd primes, then $k$ is divisible by every prime less than $n$.
I tried:
$a=p*q_0 = r_0$
$a+k=p*q_1 = r_1$
$a+(n-1)k=p*q_{n-1} = r_{n-1}$
But this is in contradiction with the red marked step:
Next the green step, $p|(a*b)\Rightarrow p|a$ v $p|b$
Is it also true like $p|(a*b) \Leftrightarrow p|a$ v $p|b$
Is this only true for prime numbers?
Yes the step in green is true, and yes it’s only true for primes. If $n$ is a composite number such that $n=pq$ then $n|pq$ obviously holds, but $n$ doesn’t divide either $p$ or $q$. For a less trivial counterexample, you can have “mismatched” factors. If $n=pq$ then $n|(pr)(qt)$ for any $r,t$ but unless $q|r$ or $p|t$ you’re not going to have $n|pr\lor n|qt$.