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?
Hint: Add $(3 - \sqrt{5})^{34}$ inside of the floor brackets and use the binomial theorem to cancel terms. Keep in mind that $0 < (3 - \sqrt{5})^{34} < 1$.
Solution:
By the binomial theorem and following the hint, we have that $\lfloor (3 + \sqrt{5})^{34} + (3 - \sqrt{5})^{34} \rfloor = \lfloor 2(3^{34} + {34 \choose 2} 3^{32} 5^1 + {34 \choose 4} 3^{30} 5^2 + \dots {34 \choose 4} 3^{4} 5^{15} + {34 \choose 2} 3^2 5^{16} + 5^{17}) \rfloor$.
The inside is an integer, so we can drop the floor brackets.
Claim: The sum of the terms not including the first two is divisible by $100$.
Proof: We find that ${34 \choose 2n}$ is even for $2 \leq n \leq 8$ by hand checking. We get a second factor of two from the $2$ multiplying it on the outside. Then all terms with at least $5^2$ as a factor and an even binomial coefficient are divisible by $100$. This deals with all of the terms except for the last two. We can see that the last two terms (without being multiplied by the external $2$) are both odd, so their sum is even. They both have at least $5^2$ as a factor so when they are multiplied by the external $2$ their sum is divisible by $100$.
So, we examine the first two terms. Note that $3^4 \equiv 81, 3^8 \equiv 61, 3^{12} \equiv 41, 3^{16} \equiv 21, 3^{20} \equiv 1 \pmod {100}$. So $3^{34} \equiv 69 \pmod {100}$ and $3^{32} \equiv 41 \pmod {100}$. So we have $2(69 + 33 * 17 * 41 * 5) \equiv 2(69 + 61 * 5) \equiv 48 \pmod {100}$
So we have that $(3 + \sqrt{5})^{34} + (3 - \sqrt{5})^{34} \equiv 48 \pmod {100}$. Since $(3 - \sqrt{5})^{34}$ is between $0$ and $1$, $(3 + \sqrt{5})^{34}$ is between $47$ and $48 \pmod {100}$, so its floor will be $\boxed{47}$.
This is a bit computation heavy, if someone can find a way that is less so I will be happy.
We use a slight modification of Michael T's approach which reduces the necessary computation and number of technical arguments . Note that $(3+\sqrt{5})^2=14+6\sqrt{5}$. Now $$\lfloor(3+\sqrt{5})^{34}\rfloor=\lfloor(14+6\sqrt{5})^{17}\rfloor=\lfloor(14+6\sqrt{5})^{17}+(14-6\sqrt{5})^{17}\rfloor-1$$ since $14-6\sqrt{5}<1$.
Now expansion with the binomial gives
$$(14+6\sqrt{5})^{17}+(14-6\sqrt{5})^{17}=2(14^{17}+{17\choose 2}14^{15}6^25+{17\choose 4}14^{13}6^45^2+\ldots+6^{16}5^8)$$
Since $6^45^2$ is a multiple of $100$, we immediately find that only the first $2$ terms matter. Note for the following that $7^4=2401\equiv1\pmod{100}$.
First term: $14^{17}2\equiv7^{17}2^{18}\equiv7^{4(4)+1}2^{9(2)}\equiv7(12)^2\equiv7(44)\equiv308\equiv8\pmod{100}$,
Second term: $2(17)(8)14^{15}6^25=10(17)(8)14^{15}6^2$, so we can get rid of a factor of $10$, and
$17(8)14^{15}6^2\equiv(7)2^37^{15}2^{15}2^23^2\equiv2^{20}7^{16}3^2\equiv4^2(1)(-1)\equiv-6\equiv4\pmod{10}$.
Hence the sum of the first two terms is $8+10*4=48 \pmod{100}$, and we conclude that $\lfloor(3+\sqrt{5})^{34}\rfloor\pmod{100}=48-1=47$.
