<s>[INST] Find all positive integers such that: $n^2-3|3n^3-5n^2+7$ 
Find all positive integers such that: $$n^2-3|3n^3-5n^2+7.$$

I did the following:
$$n^2-3|3n^3-5n^2+7,n^2-3\Rightarrow n^2-3|3n^3-5n^2+7,3n^3-9n$$
$$\Rightarrow n^2-3|5n^2-9n-7,5n^2-15\Rightarrow n^2-3|9n-8$$
Now we must have:  $9n-8\geq n^2-3$.We should determine the sign of  $n^2-9n+5$, but the equation: $n^2-9n+5=0$ does not have integer roots.How can we proceed?
 [/INST] Continuing from $\,n^2\!-\!3\mid 9n\!-\!8\ $ we can avoid quadratic inequalities and checking $n = 1,\ldots,8$
$\!\bmod n^2\!-3\!:\ \color{#c00}{n^2\equiv 3}\,\Rightarrow\, 0 \equiv \color{#c00}{n}(9\color{#c00}{n}\!-\!8)\equiv {27}\!-\!8n\,\Rightarrow\,0\equiv 9n\!-\!8+27\!-\!8n \equiv n\!+\!19\,$ hence $\,n\equiv -19\,$ so $\,0\equiv 8\!-9n\equiv 179\,$ is prime, so its divisor $\,n^2\!-\!3 = \pm1,\pm 179,\,$ so $\,n = 2.$
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