<s>[INST] Find all $n$ natural numbers such that : $n^{2}-3$ divisible by a perfect square $m>1$ Problem : 
Find all $n$ natural numbers such that : $k=n^{2}-3$ divisible by a perfect square $m>1$ 
I'm going to find the smallest number then find all this numbers 
I was tired many time of $n$ 
$n=2$ so $k=1$ $×$
$n=3$ so $k=6$ $×$ 
$n=4$ so $k=13$ $×$ 
$n=5$ so $k=22$ $×$
$n=6$ so $k=33$ $×$ 
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I don't know which number must be try it ? 
And how I find all this numbers $n=?$ 
 [/INST] $$(22+5)^2-3=22^2+220+25-3\equiv -22+22\pmod{121}$$
$$(65-4)^2-3=65^2-520+16-3\equiv-13+13\pmod{169}$$
Finding all solutions requires at least a somewhat advanced theorem, namely the law of quadratic reciprocity.
For example, there are infinitely many $n^2-3$ that are multiple of $13^2$ or $23^2$, but none that is multiple of $17^2$ or $29^2$.
If $p>3$, $n^2-3$ can be a multiple of $p^2$ iff $p\equiv \pm1\pmod {12}$. For $p=2,3$, it is not possible.
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