[Algebra]
Problem: Let $N=\left\lfloor(3+\sqrt{5})^{34}\right\rfloor$. Compute the remainder when $N$ is divided by 100 .

Solution: Let $\alpha=3+\sqrt{5}$ and $\beta=3-\sqrt{5}$, so that $N=\left\lfloor\alpha^{34}\right\rfloor$, and let $M=\alpha^{34}+\beta^{34}$. When the binomials in $M$ are expanded, terms in which $\sqrt{5}$ is raised to an odd power have opposite signs, and so cancel each other out. Therefore $M$ is an integer. Because $0<\beta<1,0<\beta^{34}<1$, and so $M-1<\alpha^{34}<M$. Therefore $M-1=N$. Note that $\alpha$ and $\beta$ are the roots of $x^{2}=6 x-4$. Therefore $\alpha^{n+2}=6 \alpha^{n+1}-4 \alpha^{n}$ and $\beta^{n+2}=6 \beta^{n+1}-4 \beta^{n}$. Hence $\alpha^{n+2}+\beta^{n+2}=$ $6\left(\alpha^{n+1}+\beta^{n+1}\right)-4\left(\alpha^{n}+\beta^{n}\right)$. Thus the sequence of numbers $\left\{\alpha^{n}+\beta^{n}\right\}$ satisfies the recurrence relation $c_{n+2}=6 c_{n+1}-4 c_{n}$. All members of the sequence are determined by the initial values $c_{0}$ and $c_{1}$, which can be computed by substituting 0 and 1 for $n$ in the expression $\alpha^{n}+\beta^{n}$, yielding $c_{0}=(3+\sqrt{5})^{0}+(3-\sqrt{5})^{0}=2$, and $c_{1}=(3+\sqrt{5})^{1}+(3-\sqrt{5})^{1}=6$. Then

$$
\begin{aligned}
& c_{2}=(3+\sqrt{5})^{2}+(3-\sqrt{5})^{2}=6 c_{1}-4 c_{0}=36-8=28 \\
& c_{3}=(3+\sqrt{5})^{3}+(3-\sqrt{5})^{3}=6 c_{2}-4 c_{1}=168-24=144
\end{aligned}
$$

and because the final result is only needed modulo 100, proceed using only remainders modulo 100.



| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $c_{n} \bmod 100$ | 6 | 28 | 44 | 52 | 36 | 8 | 4 | 92 | 36 | 48 | 44 | 72 | 56 | 48 | 64 | 92 | 96 |


| $n$ | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | 31 | 32 | 33 | 34 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $c_{n} \bmod 100$ | 8 | 64 | 52 | 56 | 28 | 44 | 52 | 36 | 8 | 4 | 92 | 36 | 48 | 44 | 72 | 56 | 48 |

Thus $N$ leaves a remainder of $48-1=\mathbf{4 7}$ when divided by 100 .
As in the previous solution, let $\alpha=3+\sqrt{5}$ and $\beta=3-\sqrt{5}$, so that $N=\alpha^{34}+\beta^{34}-1$ as argued above.

A straightforward way to compute powers of $\alpha$ and $\beta$ is by successive squaring. Paying attention to just the last two digits of the integer parts yields the following values:

$$
\begin{aligned}
\alpha^{2} & =14+6 \sqrt{5} \\
\alpha^{4} & =196+180+168 \sqrt{5} \equiv 76+68 \sqrt{5} ; \\
\alpha^{8} & \equiv 96+36 \sqrt{5} \\
\alpha^{16} & \equiv 96+12 \sqrt{5} \\
\alpha^{32} & \equiv 36+4 \sqrt{5} \\
\alpha^{34}=\alpha^{2} \cdot \alpha^{32} & \equiv 24+72 \sqrt{5} .
\end{aligned}
$$

Similarly, replacing $\sqrt{5}$ with $-\sqrt{5}$ yields $\beta^{34} \equiv 24-72 \sqrt{5}$. Thus

$$
N \equiv(24+72 \sqrt{5})+(24-72 \sqrt{5})-1 \equiv 47(\bmod 100)
$$
As in the previous solutions, let $\alpha=3+\sqrt{5}$ and $\beta=3-\sqrt{5}$, so that $N=\alpha^{34}+\beta^{34}-1$ as argued above.

Now consider the binomial expansions more carefully:

$$
\begin{aligned}
\alpha^{34} & =3^{34}+\left(\begin{array}{c}
34 \\
1
\end{array}\right) 3^{33} \sqrt{5}+\left(\begin{array}{c}
34 \\
2
\end{array}\right) 3^{32} \cdot 5+\left(\begin{array}{c}
34 \\
3
\end{array}\right) 3^{31} \cdot 5 \sqrt{5}+\cdots+\left(\begin{array}{c}
34 \\
33
\end{array}\right) 3 \cdot 5^{16} \sqrt{5}+5^{17} \\
\beta^{34} & =3^{34}-\left(\begin{array}{c}
34 \\
1
\end{array}\right) 3^{33} \sqrt{5}+\left(\begin{array}{c}
34 \\
2
\end{array}\right) 3^{32} \cdot 5-\left(\begin{array}{c}
34 \\
3
\end{array}\right) 3^{31} \cdot 5 \sqrt{5}+\cdots-\left(\begin{array}{c}
34 \\
33
\end{array}\right) 3 \cdot 5^{16} \sqrt{5}+5^{17} \\
N & =2\left(3^{34}+\left(\begin{array}{c}
34 \\
2
\end{array}\right) 3^{32} \cdot 5+\cdots+\left(\begin{array}{c}
34 \\
32
\end{array}\right) 3^{2} \cdot 5^{16}+5^{17}\right)-1 .
\end{aligned}
$$

The following argument shows that every term that is summarized by the ellipsis $(\cdots)$ in the expression for $N$ is a multiple of 50 . First, each such term has the form $\left(\begin{array}{l}34 \\ 2 k\end{array}\right) 3^{34-2 k} 5^{k}$, where $2 \leq k \leq 15$.

Thus it is enough to show that the binomial coefficient is even. Because $\left(\begin{array}{l}34 \\ 2 k\end{array}\right)=\left(\begin{array}{c}34 \\ 34-2 k\end{array}\right)$, it is enough to check this for $2 \leq k \leq 8$. Keep track of powers of 2 : $\left(\begin{array}{c}34 \\ 2\end{array}\right)$ is an integer, so



$\left(\begin{array}{c}34 \\ 4\end{array}\right)=\left(\begin{array}{c}34 \\ 2\end{array}\right) \cdot \frac{32 \cdot 31}{3 \cdot 4}$ is a multiple of $2^{3} ;\left(\begin{array}{c}34 \\ 6\end{array}\right)=\left(\begin{array}{c}34 \\ 4\end{array}\right) \cdot \frac{30 \cdot 29}{5 \cdot 6}$ is also a multiple of $2^{3} ;\left(\begin{array}{c}34 \\ 8\end{array}\right)=\left(\begin{array}{c}34 \\ 6\end{array}\right) \cdot \frac{28 \cdot 27}{7 \cdot 8}$ is a multiple of $2^{2}$; and so on.

It can also be shown that the sum of the last two terms is a multiple of 50. Again, there are plenty of factors of 5 , so it is enough to note that both terms are odd, because $\left(\begin{array}{l}34 \\ 32\end{array}\right)=\frac{34 \cdot 33}{1 \cdot 2}=$ $17 \cdot 33$.

Thanks to the initial factor of 2 in the expression for $N$ (outside the parentheses), the previous paragraphs show that $N \equiv 2\left(3^{34}+\left(\begin{array}{c}34 \\ 2\end{array}\right) 3^{32} \cdot 5\right)-1(\bmod 100)$.

Now consider the powers of 3 . Because $3^{4}=81$, we find that $3^{8}=80^{2}+2 \cdot 80+1 \equiv$ $61(\bmod 100), 3^{12} \equiv 41(\bmod 100), 3^{16} \equiv 21(\bmod 100)$, and $3^{20} \equiv 1(\bmod 100)$. (Note: those familiar with Euler's generalization of Fermat's Little Theorem will recognize this as an example, because $\phi(25)=25-5=20$.) Therefore $3^{32}=3^{20} \cdot 3^{12} \equiv 41(\bmod 100)$ and $3^{34}=3^{2} \cdot 3^{32} \equiv 69(\bmod 100)$.

Finally, $N \equiv 2(69+17 \cdot 33 \cdot 41 \cdot 5)-1 \equiv 2 \cdot 69+10 \cdot(17 \cdot 33 \cdot 41)-1 \equiv 38+10-1 \equiv \mathbf{4 7}$ $(\bmod 100)$.

Final answer: 47