If $23|3x+5y$ for some $x,y\in \mathbb{Z}$, then $23|5x-7y$? Let $x,y$ be some integers such that $23|3x+5y$. Show that $23|5x-7y$ as well.
This was my exam question and I did not solve it. Could anyone give some insight how to do this? Even after the exam I can not solve it...

High-level approach. We have the field $\mathbb{F}_{23}$ and the homomorphism $f: \mathbb{F}_{23}^2 \to \mathbb{F}_{23}$ of vector spaces defined by the formula $f(m, n) = mx + ny$, where $x$ and $y$ are fixed scalars. We know that $f(3, 5)=0$ and we want to show that $f(5, -7)=0$.
In other words, we know that $(3, 5) \in \ker f$ and we want to show that $(5,-7) \in \ker f$. To prove this, it is sufficient to show that $(5, -7)$ belongs to the linear span $\langle (3, 5) \rangle$. But this is the same as to say that $5/-7 = 3/5$ in the field $\mathbb{F}_{23}$, which is equivalent to the congruence of integers:
$$
5 \cdot 5 \equiv 3 \cdot (-7) \mod 23.
$$
This congruence is true, therefore we are done. Note how we didn't have to lucky-guess anything.
