<s>[INST] Given $a^4+1=\frac{a^2}{b^2}\left(4b^2-b^4-1\right)$, find $a^4+b^4$ 
Given $$a^4+1=\frac{a^2}{b^2}\left(4b^2-b^4-1\right)$$
  Find $$a^4+b^4$$

My approach:


*

*Verify for different values of $a$ and $b$ to see that question is correct and yes, it turns out to be $2$ always for different values of $a$ and $b$. 

*Treat the entire thing as quadratic in $a^2$, find its solution using wolfram or manually, and eventually we get the value of $a^4+b^4$ to be $2$.
Any better solution will be appreciated. 
Thank you. 
 [/INST] It is not true that we always have $a^4+b^4=2$. We can rewrite the equation as
$$
\left( a-\frac{1}{a}\right)^2+\left( b-\frac{1}{b}\right)^2=0.
$$
Now chose $a$ such that $a-1/a=i$ and $b-1/b=1$, i.e., with $a := \sqrt{ - \sqrt{3}i + 1}/\sqrt{2}$ and $b=(\sqrt{5} + 1)/2$. This comes from solving the quadratic equations $a^2-ia-1=0$ and $b^2-b-1=0$.
Then we have $i^2+1^2=0$ above satisfied and
$$
a^4+b^4=\frac{3\sqrt{5} - \sqrt{3}i + 6}{2}\neq 2.
$$
For real numbers we obtain $a^2=b^2=1$, hence $a^4+b^4=2$.
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