The Proposition: Let p be a prime with p ≡ 3 mod 4. Prove that p cannot divide both N and N + 2 for any N ∈ ℕ.
What I understand: Knowing that p = 3 mod 4, it can be said that p = 4k + 3. From this, we know p must equal an odd number.
There are also 2 cases: either n is even or n is odd.
case 1: if N is even, an even that is divisible by an odd (p) will always be even.
case 2: if N is odd, an odd that is divisible by an odd (p) will always be odd.
After identifying the base information I become confused as to where to go next with the proof. Hints don't seem to really help me out although very descriptive and well explained, step by step proof solutions do.
All help is much appreciated as this is something I want to be able to grasp better but I find myself constantly struggling with this topic.
Perhaps easier and simpler: suppose $\;p\;$ divides both $\;n\;$ and $\;n+2\;$ , then there exist integers $\;r,\,s\;$ s.t.
$$\begin{cases}I&n=rp\\{}\\ II&n+2=sp\end{cases}\;\;\stackrel{\text{subtract}\; I\; \text{from}\;\; II}\implies \;\;\;2=(s-r)p\ldots$$
Get now a contradiction...