Last two digits of $[(\sqrt{5}+2)^{2016}]$

Question

• I was trying to find the last two digits of the largest integer $\left\lfloor\left(\,\sqrt{\, 5\, }\, +\, 2\,\right)^{2016}\right\rfloor$ less than or equal to $\left(\,\sqrt{\, 5\, }\, +\, 2\,\right)^{2016}$.
• My idea is to relate $\left\lfloor\left(\,\sqrt{\, 5\, }\, +\, 2\,\right)^{2016}\right\rfloor$ to $\left\lfloor\left(\,\sqrt{\, 5\, }\, -\, 2\,\right)^{2016}\right\rfloor$, which will be the fraction part of $\vphantom{\LARGE A^{A}}\left(\,\sqrt{\, 5\, }\, +\, 2\,\right)^{2016}$. But I can't see whether it'll work.

Any idea is appretiated.

Answer 1

Note that the minimum polynomial of $\alpha =2+\sqrt 5$ is $$p(x)=x^2-4x-1$$

The other root, $\overline {\alpha}=2-\sqrt 5$, has norm less than $1$ so it makes sense to consider $$a_n= \alpha^n+ \overline {\alpha}^n$$ since for large $n$ the integer $a_n$ will be extremely close to $\alpha^n$. That sequence satisfies the recursion $$a_n=4a_{n-1}+a_{n-2}$$ with initial conditions $a_0=2, a_1=4$

You are only interested in $a_n\pmod {100}$ and a simple computation shows that the $a_n\pmod {100}$ are periodic with period $20$.

Can you finish from here?

Note: you'll need to be slightly careful and note that $\overline {\alpha}^{2016}$ is positive (and extremely small).

Answer 2

Note that $\big(r-(2+\sqrt{5})\big)\big(r-(2-\sqrt{5})\big)=r^2-4r-1$

This is the characteristic equation of $$x_{n+2}-4x_{n+1}-x_n=0$$

Since $|2-\sqrt{5}|<1$ then $|2-\sqrt{5}|^{2016}\ll 1$ thus $(2+\sqrt{5})^{2016}\approx x_{2016}$

You need to calculate $x_n\pmod{100}$.

Answer 3

As $\sqrt5-2=\dfrac1{\sqrt5+2}<1$

We actually need the last two digits of

$$f(2n)$$ $$=(\sqrt5+2)^{2n}+(\sqrt5-2)^{2n}-1 $$ $$=-1+2\sum_{r=0}^n\binom{2n}{2r}(\sqrt5)^{2n-2r}2^{2r} $$ $$=-1+2\sum_{r=0}^n\binom{2n}{2r}5^{n-r}4^r$$

$$\implies f(2016)=2\sum_{r=0}^{1008}\binom{2016}{2r}5^{1008-r}4^r$$

The last two digits essentially implies $\pmod{100}$

Now $$f(2016)\equiv2\binom{2016}05^{1008}\pmod4\equiv2\text{ as }5\equiv1\pmod2$$

and $$f(2016)\equiv2\left(\binom{2016}{2016}4^{1008}+\binom{2016}{2014}4^{1008}5\right)\pmod{5^2}\equiv2\cdot(5-1)^{1008}\equiv?$$

as $\displaystyle\binom{2016}{2014}=\binom{2016}2$ is divisible by $5$

Can you simplify and use Chinese Remainder Theorem