$n^6+18n$ is a perfect square if and only if $n=2$

Question

Let $n\in \mathbb{N}.$ Show that $n^6+18n$ is a perfect square if and only if $n=2$

My Solution:

For $n=1, n^6+18n = 19$ which is not a perfect square.

For $n=2, n^6+18n=2^6+18\cdot 2=100$ is a perfect square

We observe that for $n>2,$ $$18n<2n^3+1$$

which implies that $$n^6<n^6+18n<n^6+2n^3+1$$ $$\implies (n^3)^2<n^6+18n<(n^3+1)^2$$ $$\implies n^3<\sqrt{n^6+18n}<n^3+1$$

which implies that $\;\sqrt{n^6+18n}\;$ is a non-integer. Q.E.D

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Can anyone think of an alternative argument rather than this proof which I find very hard to motivate?

Answer 1

If $\;n^6+18n=k^2\;$ where $\;n,k\in\Bbb N\;,\;$ then

$n^3<k\;$ and $\;(k+n^3)(k-n^3)=18n\;,\;$ hence,

$2n^3\!\!\!\!\!\underset{\overbrace{\text{because}\\\;n^3<k}}{<}\!\!\!k+n^3\leqslant18n\;,\;$ consequently,

$n^2<9\;,\;$ that is, $\;\color{blue}{n=1}\;$ or $\;\color{blue}{n=2}\,.$

By inspection we get that only for $\,n=2\,,\;$ the number $\;n^6+18n\;$ is a perfect square.

Answer 2

Another way .

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We check $\thinspace n=1,2$ . Then, let $m\in\mathbb N^{+}$ and assuming that $n≥3$, we have $n^6+18n>3^4=81$, which leads to :

$$ \begin{align}&\left(n^3+\frac {9}{n^2}\right)^2-\left(\frac {9}{n^2}\right)^2=m^2\\ \implies &\left(n^5+9\right)^2-81=\left(mn^2\right)^2\\ \implies &\underbrace{\left(\color{red}{n^5}+mn^2+9\right)}_{\color{red}{>3^5}>81}\left(n^5-mn^2+9\right)=81\end{align} $$

A contradiction .