<s>[INST] Find all integers $x$ such that $\frac{2x^2-3}{3x-1}$ is an integer. Find all integers $x$ such that $\frac{2x^2-3}{3x-1}$ is an integer. 
Well if this is an integer then $3x-1 \mid 2x^2-3$ so $2x^2-3=3x-1(k)$ such that $k\in \mathbb{Z}$ from here not sure where to go I know that it has no solutions I just can't see the contradiction yet. 
 [/INST] Using the extended Euclidean algorithm in $\mathbb Q[x]$, we get
$$
25 = (-9)(2x^2-3)+(6x+2)(3x-1)
$$
Therefore, if $3x-1$ divides $2x^2-3$, then it divides $25$.
So, $3x-1 \in \{\pm 1, \pm 5, \pm 25 \}$ and there is not much to test.
An easier (but equivalent) argument is $2x^2-3 = (3x-1)q(x) + r$, where $r$ is the value of $2x^2-3$ at $x=1/3$, the root of $3x-1$. This gives $r=-25/9$. The rest follows as above.
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