Is the following proof of $(p \mid a) \land(p \not \mid b) \implies p \not \mid (a + b)$ correct?

Question

PROP Let $p$ be a prime number, and $a, b \in \mathbb{Z}$. Then we have that: $$ (p \mid a) \land(p \not \mid b) \implies p \not \mid (a + b) $$

We proceed via proof by contradiction i.e. let's assume that:

$$ (p \mid a) \land(p \not \mid b) \implies p \mid (a + b) $$

PROOF

• $p |a \iff \exists k_1 \in \mathbb{Z} : a = k_1 p$
• $p | (a + b) \iff \exists k_2 \in \mathbb{Z} : a + b = k_2 p$
• $k_2 p = a + b = k_1 p + b \iff k_2 p = k_1 p + b \iff b = (k_2 - k_1) p \iff p \mid b $
• But then we have $(p \not \mid b) \land (p \mid b)$, which is a contradiction.

$\square$

Answer 1

It's correct, but it's probably simpler to note that $p \not \mid b \Rightarrow \exists q_2, 0 \lt r \lt p \text{ such that } b=q_2p+r$ (in other words, $b$ has a positive remainder when divided by $p$), so $p \mid a \Rightarrow \exists q_1 ~(a=q_1p) \Rightarrow a+b=(q_1+q_2)p+r,$ and $0 \lt r \lt p \Rightarrow p \not \mid (a+b)$.