What are the positive real roots of $8n^8-5n^2+\sqrt{2}n+1=0$?

Question

Find all the positive roots of the equation $$8n^8-5n^2+\sqrt{2}n+1=0$$

Factoring gives $$\frac {1}{2}(2n-\sqrt {2})^2(4n^6+4\sqrt{2} n^5+6n^4+4\sqrt{2}n^3+5n^2+3\sqrt{2}n+1)$$ Hence $n>0$ is an unique root and $n=\frac{\sqrt{2}}{2}$.These calculations are very complicated and I wouldn't be able to factorise without software.Are there any easy ways to solve the problem? Or how can we simplify the calculations?

Answer 1

We combine root squaring with a search for rational roots.

Root squaring

Separate the even and odd power terms on opposite sides of the equals sign and square both sides. This gives an equation with only even powers and thus we have derived a polynomial whose roots are the aquares of the original roots.

It happens that separating the even and odd degree terms in this case also isolates the radical, so the squaring process leaves only integer coefficients.

$8n^8-5n^2+1=-\sqrt2n$

$64n^{16}-80n^{10}+16n^8+25n^4-10n^2+1=2n^2$

$64(n^2)^8-80(n^2)^5+16(n^2)^4+25(n^2)^2-12(n^2)+1=0$

Rational root search

We can search for rational roots of the above equation $n^2$ using the usual theory. Descartes' Rule of Signs allows only positive roota and the coefficients then allow only roots of the form $n^2=1/2^k$. We find that $n^2=1/2$ works. So we have a candidate positive root $n=\sqrt2/2$.

Endgame

Thus going back to the original polynomial $8n^8-5n^2+\sqrt2n+1$ we test the factor $n-(\sqrt2/2)$. Upon dividing we find that this factor holds and has multiplicity $2$, giving the factorization quoted in the question.

Answer 2

You observe that, we have :

$$ \begin{align}5&\thinspace=\thinspace\frac{8n^8+n\sqrt 2+1}{n^2}\\ &\thinspace=\thinspace 8n^6+\color{#c00}{\frac{\sqrt 2}{n}}+\color{#0a0}{\frac {1}{n^2}}\end{align} $$

The positivity of $n$ and the uniquity of the positive root of the polynomial give us an idea of ​​the applicability of the AM-GM inequality. Thus, we will try to apply the AM-GM by rewriting the polynomial as follows :

$$ \small{\begin{align}5&\thinspace=\thinspace8n^6+\color{#c00}{\frac {\sqrt 2}{2n}+\frac {\sqrt 2}{2n}}+\color{#0a0}{\frac {1}{2n^2}+\frac {1}{2n^2}}\\&\thinspace≥\thinspace5\sqrt [5]{8n^6\cdot\frac {\sqrt 2}{2n}\cdot\frac {\sqrt 2}{2n}\cdot\frac {1}{2n^2}\cdot\frac {1}{2n^2}}\\ &\thinspace=\thinspace5\end{align}} $$

As you know, the equality holds iff, when :

$$ \begin{align}&8n^6=\frac {\sqrt 2}{2n}=\frac {1}{2n^2}\\ \implies &n=\frac {1}{\sqrt 2}=\frac {\sqrt 2}{2}\end{align} $$

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Finally, we observe that :

$$8n^8-5n^2+n\sqrt 2+1≥0$$

holds $\forall n\in\mathbb R^{+}$ and the equality occurs iff, when $n=\dfrac {\sqrt 2}{2}$ .

Answer 3

One approach that suggests a root to try is to look at the Newton polygon for each $\mathbb{Q}_p(\sqrt{2})$. It happens that every Newton polygon is a flat line except at $p=2$ where we have up to six roots of $2$-adic valuation $-\frac{1}{2}$ which suggests trying $\frac{1}{\sqrt{2}}$ or $\frac{-1}{\sqrt{2}}$.

The reason to suspect such an approach could work at all is the fact that we can imagine $\mathbb{Q}(\sqrt{2})$ contained within all the $\mathbb{Q}_p(\sqrt{2})$ and $\mathbb{R}$ with their absolute values related by a product formula, $$1=\prod_{p \le \infty}|x|_p$$