For what integer values of $y$ is $\frac{3y-1}{y-3}$ an integer? I have encountered this as part of a bigger problem but I really don't know how to go on about it. I would also appreciate it if you could specify a certain technique to follow when facing such a problem.
 A: Write it as $$\frac{3y-1}{y-3}=\frac { 3\left( y-3 \right) +8 }{ y-3 } =3+\frac { 8 }{ y-3 } $$
A: write $$\frac{3y-1}{y-3}=\frac{3y-9+8}{y-3}=3+\frac{8}{y-3}$$
A: The technique is by using modulos. $$3y-1=3(y-3)+8 \equiv 0 \pmod {y-3}$$
So $8 \equiv 0 \pmod{y-3}$. I think you can continue from here. 
A: Please, note that we encounter here with a rational function of the form $x = f(y) = \frac{g(y)}{h(y)}$ where $g(y) = 3y - 1$ and $h(y) = y - 3$ with a restriction that $h(y) \neq 0 \iff y \neq 3$.
For any rational function it is possible to express $f(y)$ in a form of $\frac{a}{y - p} + q$ where $a, p, q \in \mathbb{R}\ \wedge\ a, p, q= \textrm{const}$ and lines $y = p$ and $x = q$ are the asymptots of the function $f(x)$.
Hence,
$$f(y) = \frac{3y - 1}{y - 3} = \frac{3(y - 3) + 8}{y - 3} = \frac{8}{y - 3} + 3$$
This yields, that
$$f(y) \in \mathbb{Z} \iff y - 3\ \textrm{mod}\ 8 \equiv 0$$
In other words, $f(y)$ is an integer if and only if $8$ is divisible by $y-3$, which implies that $8 = k(y - 3)$ for some $k \in \mathbb{Z}$ ($8$ is a multiple of $y-3$).
Finally,
$$8 = k(y - 3)$$
$$y = \frac{8}{k} + 3$$
This is satisfied for all $k \in \{z \in \mathbb{Z}\backslash \{0\}\ |\ 8\ \textrm{mod}\ z \equiv 0\} = \{1,-1,2,-2,4,-4,-8,8\}$ since $y$ must be an integer.
Therefore solution set can be given as
$$y \in \left\{\frac{8}{k} + 3\ |\ k \in \{1,-1,2,-2,4,-4,-8,8\}\right\}$$
A: hint: Suppose it equals an integer $n$ and rewrite it as $(y-3)(n-3)=8$. This leaves few possibilities to go through.
A: Hint $\ y\!-\!3\mid f(y)\iff y\!-\!3\mid f(3)\ $ for any polynomial $f$ with integer coefficients, because
$\ {\rm mod}\,\ y\!-\!3\!:\ y\equiv 3\,\Rightarrow\, f(y)\equiv f(3)\ $ by the Polynomial Congruence Rule.
Equivalently $\, f(y)\equiv f(n)\pmod{y\!-\!n},\ $ the Polynomial Remainder Theorem.
Alternatively we can apply the  Euclidean algorithm and the remainder theorem as follows
$$\gcd(y\!-\!3,f(y))\, =\, \gcd(y\!-\!3,\,f(y)\bmod y\!-\!3)\, =\, \gcd(y\!-\!3,f(3))$$
