Does $\lfloor(4+\sqrt{11})^{n}\rfloor \pmod {100}$ repeat every $20$ cycles of $n$?

I recently came across a post on SO, asking to calculate the least two decimal digits of the integer part of $(4+\sqrt{11})^{n}$, for any integer $n \geq 2$ (see here).

The author presented a Java implementation using a BigInteger class and so forth, but the answer (given by someone else) was much simpler:

$$\forall n \geq 2 : \lfloor(4+\sqrt{11})^{n}\rfloor \pmod{ 100}=\lfloor(4+\sqrt{11})^{n+20}\rfloor \pmod {100}.$$

So in essence, we only need to calculate $(4+\sqrt{11})^{n}$ for every $n$ between $2$ and $21$.

I have unsuccessfully attempted to find a counterexample, using a BigRational class and a fast-converging $n$th-root algorithm for calculating $\sqrt[n]{A}$.

My question is then, how can we prove or refute the conjecture above?

Pardon my tags on this question, wasn’t sure what else to put besides irrational-numbers.

Solutions Collecting From Web of "Does $\lfloor(4+\sqrt{11})^{n}\rfloor \pmod {100}$ repeat every $20$ cycles of $n$?"

Consider the linear recurrence relation $a_{n+2} = 8a_{n+1} – 5a_n$ with $a_0 = 2$ and $a_1 = 8$. It is clear that $a_n$ is an integer for all $n \geq 0$. Also, this recurrence has the following closed-form solution for all $n \geq 0$:

$a_n = (4+ \sqrt{11})^n + (4 – \sqrt{11})^n$

Since $0 < 4 – \sqrt{11} < 1$, it follows that $\lfloor(4+ \sqrt{11})^n\rfloor = a_n – 1$ for all $n \geq 0$. Periodicity of $\lfloor(4+ \sqrt{11})^n\rfloor$ follows almost immediately now: first we check that $54 = a_{2} \equiv a_{22} \text{ mod } 100$ [Edit: and $92 \equiv a_{3} \equiv a_{23} \text{ mod } 100$]. Since $a_n$ is defined via a linear recurrence, it follows that $a_n \equiv a_{n+20} \text{ mod } 100$ for all $n \geq 2$, so $\lfloor(4+ \sqrt{11})^n\rfloor \equiv \lfloor(4+ \sqrt{11})^{n+20}\rfloor \text{ mod } 100$ for all $n \geq 2$ as well.