$(\sqrt{10}+3)^{2010}$ What are the first 100 digits of $(\sqrt{10}+3)^{2010}$ after the decimal point? 
I used a calculator to figure out $(\sqrt{10}+3)^{2}$ up to of $(\sqrt{10}+3)^{6}$ and they all have increasing numbers of nines after the decimal point. Does that pattern continue, and if so, why?
 A: Let $a = 3 + \sqrt{10}$ and $b = 3 - \sqrt{10}$.  Note that $ab = 3^2 - 10 = -1$.  Now $a^n + b^n$ is a positive integer, in fact $c_n = a^n + b^n$ satisfies the recurrence $c(n) = 6 c(n-1) + c(n-2)$ because $a^n$ and $b^n$ do,
with initial conditions $c(0)=2$, $c(1) = 6$. Thus 
$$ a^{2010} = c(2010) - b^{2010}$$
where $c(2010)$ is a large positive integer while $b^{2010}$ is a small positive number.  In fact $|b| < 10^{-0.7897}$, so $b^{2010} < 10^{-1587}$, i.e.
$a^{2010}$ has at least $1587$ nines after the decimal point.
A: When $a, b, n$ are positive whole numbers we have the following sum of conjugate quadratic surds
$(a+\sqrt{b})^n+(a-\sqrt{b})^n$=exactly a whole number
There are several ways to prove this. One way is to expand both powers with the binomial theorem.  When you add them together, the terms containing odd powers of $\sqrt{b}$ cancel out and the terms containing even powers of $\sqrt{b}$ which are whole numbers doubled up (so the sum is actually an even whole number).
Here with $a=3,b=10,n=2010$ the conjugate term $(a-\sqrt{b})^n$ is very small  (because the base is absolutely much smaller than $1$ and the exponent is large) and positive (because the exponent is even).  You are to check whether it's so small that the other term in the sum, which is what you want, has to have 100 or more nines after the decimal point to complete a whole number sum.
