<s>[INST] Alternative way to calculate the sequence Given $a>0$ that satisfies $4a^2+\sqrt 2 a-\sqrt 2=0$.
Calculate $S=\frac{a+1}{\sqrt{a^4+a+1}-a^2}$.
Attempt:
There is only one number $a>0$ that satisfies $4a^2+\sqrt 2 a-\sqrt 2=0$, that is
$a=\frac{-\sqrt{2}+\sqrt{\Delta }}{2\times 4}=\frac{-\sqrt{2}+\sqrt{2+16\sqrt{2}}}{8}$
However obviously if you replace it directly to calculate $S$, it would be extremely time-consuming.
Is there another way? I think the first equation can be changed (without needing to solve it) to calculate $S$, as when I use brute force (with calculator), I found out that $S\approx 1.414213562\approx\sqrt 2$.
 [/INST] Rewrite using the value of $a^2 = \frac{1-a}{2\sqrt2}$
$$\frac{a+1}{\sqrt{\tfrac{(a-1)^2}{8}+a+1}+\tfrac{(a-1)}{2\sqrt{2}}} = \frac{(a+1)2\sqrt2}{\sqrt{a^2+6a+9}  + a-1} = \frac{(a+1)2 \sqrt 2}{(a+3) + a-1} \\= \sqrt{2}$$
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