<s>[INST] How find the sum of the last two digits of $(x^{2})^{2013} + \frac{1}{(x^{2})^{2013}}$ for $x + \frac{1}{x} = 3$? Let x be a real number so that $x + \frac{1}{x} = 3$. How find the sum of the last two digits of $(x^{2})^{2013} + \frac{1}{(x^{2})^{2013}}$?
 [/INST] Square your equation to get:
$$x^2+\frac{1}{x^2}=3^2 - 2$$
And again:
$$x^4+\frac{1}{x^4}=(3^2 - 2)^2-2$$
So:
$$x^{2^n}+\frac{1}{x^{2^n}}=(\ldots((3^2 - 2)^2-2)^2 -2)\ldots)^2 -2$$
Define:
$$a_{n+1} = a_n^2 - 2$$
Now let $a_n = 100 b_n + y_n$. So that:$$a_{n+1} = 100^2 +200 b_n y_n+ y_n^2 - 2$$
$$y_{n+1} = y_n^2 - 2\mod 100$$
Start with $y_1=7$. $y_2 = 47$ and $y_3 = 107 \equiv 7$.
Edit:
This works to get us a answer if we had $2048$ instead of $2013$. 
Working on an alternate solution.
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