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`

.