Finding the correct statements about $(5+2\sqrt{6})^{2n+1} = S + t$ with $S$ integer and $0 \leq t < 1$ Problem:

If $n$ is a positive integer and $(5+2\sqrt{6})^{2n+1} = S + t$, where $S$ is an integer and $0 \leq t < 1$, then
(a) $S$ is an odd integer
(b) $S + 1$ is not divisible by $9$
(c) The integer next above $(5+2\sqrt{6})^{2n+1}$ is divisible by $10$
(d) $S-1 = \frac{t}{t-1}$

Please guide me how to proceed with this question...

Everything flows from the hint by Ross Millikan. 
Let $(5+2\sqrt{3})^{2n+1}=a_n+b_n\sqrt{3}$. (we can imagine computing $a_n$ and $b_n$ by expanding using the Binomial Theorem). Note that $a_n$ and $b_n$ are integers. Now imagine expanding $(5-2\sqrt{3})^{2n+1}$. We get $a_n-b_n\sqrt{3}$. It follows that
$$(5+2\sqrt{3})^{2n+1}+(5-2\sqrt{3})^{2n+1}=2a_n.\tag{$1$}$$ 
Define $S_n$ and $t_n$ as in the question, except we have added subscripts to make clear the dependence on $n$. By the hint of Ross Millikan, $S_n=2a_n-1$.  We can conclude immediately that $S$ is odd. 
Note that
$$\begin{align}a_{n+1}+b_{n+1}\sqrt{3}&=(5+2\sqrt{3})^{2n+3}=(a_n+b_n\sqrt{3})(5+2\sqrt{3})^2\\  &= 37a_n+60b_n+(20a_n+37b_n)\sqrt{3}\end{align}.\tag{$2$}$$
Thus
$$a_{n+1}=37a_n+60b_n\qquad\text{and}\qquad b_{n+1}=20a_n +37b_n.\tag{$3$}$$
Note that $a_0=5$ and $b_0=2$. The integer just above $S_n$ is $2a_n$. This is divisible by $10$ if $n=0$. It follows by induction, using the first recurrence in $(3)$, that $10$ divides $2a_n$, that is, $S_n+1$, for all $n$. 
About non-divisibility of $S_n+1=2a_n$ by $9$, we prove that $a_n$ is not divisible by $3$. Certainly $a_0=5$, so $a_0$ is not a multiple of $3$. But then by the recurrence $a_{n+1}=37a_n+60b_n$, we see that $a_1$ is not divisible by $3$, but then again by the recurrence, $a_2$ is not divisible by $3$, and so on. 
The computation in (d) is left to you. 
