Evaluating $\lfloor (3 + \sqrt{5})^{34} \rfloor \pmod {100}$ The problem is to evaluate $\lfloor (3 + \sqrt{5})^{34} \rfloor \pmod {100}$
No calculators are allowed. 
I think I have to get rid of $\sqrt{5}$ somehow since it is irrational and would make it hard to find the floor without doing it numerically. But I'm not sure how.

This answer inspired by the Fibonacci sequence, in that $$F_n = \frac{\varphi^n - \overline{\varphi}\,^n}{\sqrt 5} = \frac{\lfloor{\varphi^n}\rfloor + 1}{\sqrt 5} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n\,_{2,1}$$
which gives us an easy way to compute $\lfloor \varphi^n \rfloor$ using the above matrix.  So attempting to find a matrix that allows the computation of $\lfloor\phi^n\rfloor$ for $\phi = 3 + \sqrt 5$ and $\overline \phi = 3 - \sqrt 5$, consider the matrix:
$$M = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \tag 1$$
It has characteristic equation 
$$P(x) = x^2 - (D+A)x + AD - BC \tag 2$$
which to have the desired eigenvalues, the characteristic equation must also be:
$$P(x) = (x - \phi)(x - \overline \phi) = x^2 - 6x + 4 \tag 3$$
Eliminating variables $C$ and $D$ between (1) (2) and (3), it leaves:
$$M = \begin{bmatrix} A & B \\ \frac{(6-A)A - 4}{B} & 6 - A \end{bmatrix} \tag 4$$
Now any choice of $A$ and $B$ will work, but to choose something easy, consider $A = 3$ to make the diagonal nice and $B=1$ to make the rest nice:
$$M = \begin{bmatrix} 3 & 1 \\ 5 & 3 \end{bmatrix}$$
with the eigenvalue decomposition being:
$$M = 
\begin{bmatrix} 1 & 1 \\ \sqrt{5} & -\sqrt{5}\end{bmatrix}
\begin{bmatrix} 3 + \sqrt{5} & 1 \\ 0 & 3 - \sqrt{5}\end{bmatrix}
\begin{bmatrix} 1 & 1 \\ \sqrt{5} & -\sqrt{5}\end{bmatrix}^{-1}
$$
and it follows:
$$M^n =
\begin{bmatrix} 1 & 1 \\ \sqrt{5} & -\sqrt{5}\end{bmatrix}
\begin{bmatrix} \phi^n & 1 \\ 0 & \overline \phi\,^n\end{bmatrix}
\begin{bmatrix} 1 & 1 \\ \sqrt{5} & -\sqrt{5}\end{bmatrix}^{-1}
=
\begin{bmatrix}
   \frac{\phi^n + \overline \phi\,^n}{2} & \frac{\phi^n + \overline \phi\,^n}{2\sqrt{5}} \\
   \frac{\phi^n + \overline \phi\,^n}{2 / \sqrt{5}} & \frac{\phi^n + \overline \phi\,^n}{2} \\
\end{bmatrix}
$$
So we have $2 M^n\,_{1,1} = \phi^n + \overline \phi\,^n = \lfloor \phi^n \rfloor + 1$, altogether:
$$\lfloor(3 + \sqrt{5})^n\rfloor = 2 {\begin{bmatrix} 3 & 1 \\ 5 & 3 \end{bmatrix}^n}_{1,1} - 1$$
$$\downarrow$$
$$\boxed{\lfloor(3 + \sqrt{5})^n\rfloor \equiv 2 {\begin{bmatrix} 3 & 1 \\ 5 & 3 \end{bmatrix}^n}_{1,1} - 1 \pmod{100}} \tag{!!}$$
Matrix modular exponentiation with integers is fairly easy to do with repeated squaring:
$$
\begin{array} {|c|c|} \hline
M^1                  & \begin{bmatrix} 3 & 1 \\ 5 & 3 \end{bmatrix} \\ \hline
M^2 = (M^1)^2        & \begin{bmatrix} 14 & 6 \\ 30 & 14 \end{bmatrix} \\ \hline
M^4 = (M^2)^2        & \begin{bmatrix} 76 & 68 \\ 40 & 76 \end{bmatrix} \\ \hline
M^8 = (M^4)^2        & \begin{bmatrix} 96 & 36 \\ 80 & 96 \end{bmatrix} \\ \hline
M^{16} = (M^8)^2       & \begin{bmatrix} 96 & 12 \\ 60 & 96 \end{bmatrix} \\ \hline
M^{32} = (M^{16})^2    & \begin{bmatrix} 36 & 4 \\ 30 & 36 \end{bmatrix} \\ \hline
M^{34} = M^{32}M^2     & \begin{bmatrix} 24 & 70 \\ 60 & 24 \end{bmatrix} \\ \hline
\end{array}$$
So $\lfloor(3 + \sqrt{5})^{34}\rfloor \equiv 2 \times 24 - 1 \equiv 47 \pmod{100}$

It seems that using the above style argument, one can say in general that if $0 < X - \sqrt{Y} < 1$, then the follow identity holds:
$$\lfloor (X + \sqrt{Y})^n \rfloor = 2{\begin{bmatrix} X & Y \\ 1 & X \end{bmatrix}^n}_{1,1} - 1$$
I wonder if this is a well known identity?
