veronica1999 said:
If Rn = 1/2(a^n + b^n) where a= 3+ 2root2 b= 3 - 2root2 and n= 0,1,2,3,4,5,...
then R12345 is an integer. Its units digit is..
I didn't have any ideas so I started to plug in numbers looking for a pattern.
I am sure this is not the right approach.
R0 = 1
R1 = 3
R2= 7
R3= 9
Could I get some help?
The following might help although it requires some computation.
Note that $\alpha=3+2\sqrt{2}$ and $\beta=3-2\sqrt{2}$ are roots of $x^2-6x+1=0$.
That is, they are roots of $x^3-6x^2+x=0$. And again, this means they are also roots of $(x^{4115})^3-6(x^{4115})^2+x^{4115}=0$.
The above when simplified gives $x^{12345}=6x^{8230}-x^{4115}$.
Substituting $\alpha$ and $\beta$ in the above and adding we get:
$(3+2\sqrt{2})^{12345}+(3-2\sqrt{2})^{12345}=6((3+2\sqrt{2})^{8230}+(3-2\sqrt{2})^{8230})-((3+2\sqrt{2})^{4115}+(3-2\sqrt{2})^{4115})$
Basically, we now have
lower exponents for which we have to calculate. Also, we just need the remainder with $10$. The actual number is really HUGE. If you are familiar with the chinese remainder theorem then you can first find the remainder mod $5$ and then remainder mod $2$ to finally find remainder mod $10$. While finding remainder mod $5$, you can knock the hanging $6$ out. I am not sure if this will solve the question but still.
Another approach which comes to mind is the use of the binomial theorem.
To simplyfy computation observe $3+2\sqrt{2}=(1+\sqrt{2})^2$ and $3-2\sqrt{2}=(1-\sqrt{2})^2$. I was not able to find remainders mod $10$ of the binomial coefficients which come in our way though.
I have to go to class, I will get back to this later. If you are able to solve this using the above mentioned things or otherwise then post it here.
