I want to prove this statement: Let x and y be integers. If $d\in \mathbb{Z}$ such that $d|(ax+by)$ $\forall a, b\in \mathbb{Z},$ then $d|x$ and $d|y$.
I thought of rewriting the statement logically and then negating it and finding a contradiction in the negation, but I apparently kept on getting that the negation was true, which makes no sense.
Now, I'm thinking of saying, "Assume that d divides (ax+by) for all integers a and b. In particular, for $a =1$ and $b=0$, $d|x$. For $a=0$ and $b=1$, d|y. Thus, if $d|(ax+by)\forall a,b\in \mathbb{Z}$, then $d|x \wedge d|y$. Is this okay for a universal quantifier?
This answer is based on the original version of the question text which didn't include any method to prove the statement.
You have that $d\in \mathbb{Z}$ such that $d|(ax+by)$ $\forall a, b\in \mathbb{Z}$. Choose $a = 1, b = 0$ to get that $d \mid x$, then next choose $a = 0, b = 1$ to get that $d \mid y$. This shows that $d \mid x$ and $d \mid y$.
Although you don't ask for it, note the converse is also true. In particular, if $d \mid x$ and $d \mid y$, then $d$ will divide all linear combinations of $x$ and $y$, i.e., $d|(ax+by)$ $\forall a, b\in \mathbb{Z}$.