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Integrating squared absolute value of a complex sequence

$$\sqrt{(5+2\sqrt6)^x}+\sqrt{(5-2\sqrt6)^x}=10$$

So I have squared both sides and got:

$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2\sqrt{1^x}=100$$

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$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2=100$$

I don’t know what to do now

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You don’t have to square the equation in the first place.

Let $y = \sqrt{(5+2\sqrt{6})^x}$, then $\frac{1}{y} = \sqrt{(5-2\sqrt{6})^x}$. Hence you have $y + \frac{1}{y} = 10$ i.e. $y^2 + 1 = 10y$ i.e. $y^2-10y+1 = 0$.

Hence, $(y-5)^2 =24 \Rightarrow y = 5 \pm 2 \sqrt{6}$.

Hence, $$\sqrt{(5+2\sqrt{6})^x} = 5 \pm 2\sqrt{6} \Rightarrow x = \pm 2$$

(If you plug in $x = \pm 2$, you will get $5+2\sqrt{6} + 5-2\sqrt{6} $ which is nothing but $10$)

You’ve already seen that $(5-2\sqrt{6})(5+2\sqrt{6})=1$ when you squared both sides. This means that $5-2\sqrt{6}=\frac{1}{5+2\sqrt{6}}$, so your last equation can be rewritten as $$\left(\frac{1}{5+2\sqrt6}\right)^x+(5+2\sqrt6)^x+2=100$$

or, letting $y=(5+2\sqrt{6})^x$,

$$\frac{1}{y}+y+2=100$$

so

$$1+y^2+2y=100y$$

which is quadratic in $y$. Solve this for $y$, then use that solution to solve for $x$.

While this problem succumbs to high-school algebra, with a little college algebra one can go much further to derive recurrences and addition formula for the power sums $\rm\ s_n\: =\ w^n + w’^n\ $ of the roots of an arbitrary quadratic polynomial $\rm\ f(x)\ =\ (x-w)\ (x-w’)\ =\ x^2 – b\ x + c\:.\ $ Namely, we have $\rm\ s_{n+1}\ =\ b\ s_n – c\ s_{n-1}\ $ and, more generally, putting the recurrence in matrix form (e.g. see the Fibonacci case) yields the addition formula $\rm\ s_{m+n} = s_m\ s_n – c\ s_{m-n}\:.\: $ This enables very rapid computation of the sequence by what amounts to repeated squaring of the matrix representing the shift operator (e.g. see the cited Fibonacci case). For example we obtain a doubling formula via $\rm\:m=n\:$ in the addition formula: $\rm\ s_{\:2\:n}\ =\ s_n^2 – c\ s_0\ =\ s_n^2 – 2\:.\:$ In the example at hand we have $\rm\ s_0,\ s_1,\:\ldots\ =\ 2,10,98,970,9602,95050,940898\ =\ s_6\ $ and, indeed, $\rm\ s_6\ =\ s_3^2 – 2\ =\ 970^2-2\:.\ $ This is a special case of general results about Lucas-Lehmer sequences. For further discussion see Ribenboim: *The New Book of Prime Number Records*.

Observe that $(5 – 2\sqrt{6}) = \frac{1}{5+2\sqrt{6}}$.

So, if we set $a = 5+ 2 \sqrt{6}$, we have

\begin{equation}

a^x +a^{-x} +2 = 100.

\end{equation}

The above expression is symmetric in $x$, so if $x = k$ is a solution, $x = -k$ is also a solution. Consider $f(x) = a^x +a^{-x} +2$ for $x>0$. You can show that this is an increasing function for $x > 0$. This implies that there can be at most one value of $x>0$ for which the equality is satisfied. You can check by substitution that $x = 2$ is a solution. So, the only solutions for the equation are $x = +2$ and $x=-2$.

Rather strangely it is possible to spot a solution from the equation simply by realising $$\left(5+2\sqrt6\right)+\left(5-2\sqrt6\right)=10$$

and once you know $5+2\sqrt{6} = \frac{1}{5-2\sqrt{6}}$ then you can spot the other.

$(\sqrt{(5+2\sqrt6)^x}+(\sqrt{(5-2\sqrt6)^x}=10$

We have, $5+2\sqrt6 = (\sqrt{2} + \sqrt{3})^2$ and $5 – 2\sqrt6 = (\sqrt3-\sqrt2)^2$

The equation $\Leftrightarrow \left(\sqrt{2} + \sqrt{3}\right)^x + \left(\sqrt3-\sqrt2\right)^x = 10$

Let $$t = \left(\sqrt3+\sqrt2\right)^x = \frac{1}{\left(\sqrt3-\sqrt2\right)^x}$$ with $t\neq 0$

The equation $\Leftrightarrow t + \frac1t – 10 = 0 \Leftrightarrow t^2 – 10t + 1 = 0 $

$\Leftrightarrow t = 5 \pm 2\sqrt6$

With case $t = 5 + 2\sqrt6 \Rightarrow \left(\sqrt3+\sqrt2\right)^x = 5 + 2\sqrt6 \Rightarrow x = \log_{\sqrt3+\sqrt2}(5+2\sqrt6) = 2$

With case $t = 5 – 2\sqrt6 \Rightarrow \left(\sqrt3+\sqrt2\right)^x = 5 – 2\sqrt6 \Rightarrow x = \log_{\sqrt3+\sqrt2}(5-2\sqrt6) = -2$.

Done!

Wolfram Alpha can help:

Equation

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