Suppose $p,~q$ are two prime numbers such that $p^2-p<q<2p^2$ and $p\geq 5$. I want to prove that $q>4p$ and $i>s-1$, where $i,~s$ are integers such that $q=ip+s$ where $1\leq s<p~$ i.e. $i,~s$.
My attempt: $q>4p$ since $q>p^2-p=p(p-1)$ and $p\geq 5$ i.e. $p-1\geq 4$.
Now i can't reach to the relation between $i$ and $s$.
Can you prove it, please?
Thanks in advance.
Since $$p^2-p\geq 4p\implies p(p-5)\geq0$$
is true for all $p\geq 5$ we are almost finish.
Say $i\leq s-1 $ then from $q=ip+s$ and $s\leq p-1$ we get $$q\leq (s-1)p+s \leq p(p-2)+p-1 = p^2-p-1$$
which is a contradiction since $q>p^2-p$.