Question: Prove $x,y\in\mathbb{Z},\Bigl((5\nmid xy)\to(5\nmid x\land 5\nmid y)\Bigr)$ where $\forall a,b\in\mathbb{Z},\bigl((a\nmid b)\leftrightarrow(\forall k\in\mathbb{Z},b\neq ak)\bigr)$ and $\forall a,b\in\mathbb{Z},\bigl((a\mid b)\leftrightarrow(\exists k\in\mathbb{Z},b=ak)\bigr)$.
So my question is, what in the world is this problem asking and how do I begin to solve it? I know I have to use either proof by contradiction or contrapositive, but I'm struggling so much with this question. Could someone please help me?
This problem is, ultimately, asking you to prove a statement of the form $$ (c\mid x\lor c\mid y)\to c\mid xy $$ where $c=5$. Why the problem is so poorly posed I do not know. As stated, this is what you have to prove: $$ (\forall x,y\in\mathbb{Z})(5\nmid xy\to(5\nmid x\land 5\nmid y)).\tag{1} $$ We may rewrite $(1)$ into an equivalent statement using the contrapositive: $$ (\exists x,y\in\mathbb{Z})((5\mid x\lor 5\mid y)\to 5\mid xy).\tag{2} $$ If $5\mid x$, then $x=5\ell$ for some $\ell\in\mathbb{Z}$. Also, if $5\mid y$, then $y=5m$ for some $m\in\mathbb{Z}$. Since we have an "or" statement, we have essentially three possibilities: $5\mid x$, or $5\mid y$, or $5\mid x$ and $5\mid y$. This translates, respectively, to $xy=5\ell y$, or $xy=5m x$, or $xy=5(5m\ell)$. As you can see, $5\mid xy$ for all possibilities, hence proving your result (this proves your result because we just proved $(2)$ which is equivalent to $(1)$).