Find all natural numbers $n$ such that $2n+1$ divides $n^{4}+n^{2}$.

Question

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.

Answer 1

Yes, your solution is correct.

You can also avoid modular arithmetic. It is correct that $n^2$ and $2n+1$ are relatively prime so $$2n+1\mid n^2+1$$ Since $2n+1\mid 4n^2-1$ we have $$2n+1\mid 4(n^2+1)-(4n^2-1) =5$$ and you are done.

Answer 2

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.

Answer 3

Correct. Simpler: $\bmod 2n\!+\!1\!:\,\ \color{#c00}{2n\equiv -1}\,\Rightarrow\, 2^{\large 4}(n^{\large 4}+n^{\large 2})\equiv (\color{#c00}{2n})^{\large 4}+4(\color{#c00}{2n})^{\large 2} \equiv 1+4$