<s>[INST] Find all natural numbers $n$ such that $2n+1$ divides $n^{4}+n^{2}$. 
Find all natural numbers $n$ such that $2n+1$ divides $n^{4}+n^{2}$.

My attempt:-
  We need to find natural numbers, $n$, for which $2n+1\mid n^{4}+n^{2}$. We write, $$n^{4}+n^{2}=n^{2}(n^{2}+1)$$. It can be easily proved that $(n^{2},2n+1)=1$(so I think I need not type it here because that will increase the length of this post).This implies that, $$n^{2}+1\equiv 0\pmod {2n+1}\implies 4n^2+4\equiv 0\pmod{2n+1} \implies (2n+1)^{2}+3\equiv 4n\pmod{2n+1} \implies 4n\equiv 3\pmod{2n+1} \implies 2(2n+1)\equiv 5\pmod{2n+1}$$. 
Hence, $2n+1\mid 5$  and this implies $n=2$, which is the only solution. Does this look good? I think this solution is correct but I'm not satisfied with it, because it took me quite a long time before I came up with this argument. Can this be solved without any such manipulations? Thank you.
 [/INST] Divide $$ \frac{n^4 + n^2}{2n+1} = \frac{n^3}{2} - \frac{n^2}{4} + \frac{5n}{8} - \frac{5}{16} + \frac{5}{8 (2n+1)}$$ and note that the $\frac{5}{8 (2n+1)}$ term, if rewritten in lowest terms, will have an odd factor in the denominator unless $n = 2$, so the whole expression cannot be an integer. Then all that remains is checking that $n + 2$ is in fact a solution.
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