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Prove that: $$\dfrac{1}{2^{180}a^{360}}\dfrac{(a^{720}-1)(a^2-1)}{a^{2}+1} = \dfrac{\left(1+\dfrac{\sqrt{3}}{2}\right)^{180} - \left(1-\dfrac{\sqrt{3}}{2}\right)^{180}}{\sqrt{3}}$$ where: $$a = \dfrac{1+\sqrt{3}}{\sqrt{2}}.$$

This seemed like a very challenging question but the fact that we have a telescoping binomial sum in the numerator of the LHS helps. I think if we can simplify the LHS sufficiently we might be able to prove it by just equating both sides of the equation.

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  • $\begingroup$ Possible hint: $a = 2 \cos \frac \pi{12}$. :) $\endgroup$
    – Kaster
    Commented Mar 20, 2016 at 3:37

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HINT:

Observe that $a^2=2+\sqrt3$

$\implies1+\dfrac{\sqrt3}2=\dfrac{a^2}2$ and $1-\dfrac{\sqrt3}2=\dfrac1{2a^2}$

Use Componendo and Dividendo to find $\dfrac{a^2-1}{a^2+1}=?$

Now

$$\dfrac{1}{2^{180}a^{360}}\dfrac{(a^{720}-1)(a^2-1)}{a^{2}+1}=\left[\left(\dfrac{a^2}2\right)^{180}-\left(\dfrac1{2a^2}\right)^{180}\right]\cdot\dfrac{a^2-1}{a^2+1}$$

Can you reach home from here?

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  • $\begingroup$ How do I use this fact? $\endgroup$
    – user19405892
    Commented Mar 20, 2016 at 3:29
  • $\begingroup$ @user19405892, Updated $\endgroup$
    – lab bhattacharjee
    Commented Mar 20, 2016 at 3:29
  • $\begingroup$ Can you explain the compenendo part? $\endgroup$
    – user19405892
    Commented Mar 20, 2016 at 3:36
  • $\begingroup$ @user19405892, $$\dfrac{1}{2^{180}a^{360}}(a^{720}-1)=\dfrac1{2^{180}}\left(a^{360}-\dfrac1{a^{360}}\right)=\dfrac1{2^{180}}\left((a^2)^{180}-\dfrac1{(a^2)^{180}}\right)$$ $\endgroup$
    – lab bhattacharjee
    Commented Mar 20, 2016 at 3:37
  • $\begingroup$ Impressive algebra! $\endgroup$
    – NoChance
    Commented Mar 20, 2016 at 3:39

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