MHB Discovering the Unit Digit of Rn: Solving for n = 0 to 5

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The discussion centers on finding the units digit of Rn, defined as Rn = 1/2(a^n + b^n) with a = 3 + 2√2 and b = 3 - 2√2, for n = 0 to 5. Initial calculations show R0 = 1, R1 = 3, R2 = 7, and R3 = 9, leading to the discovery that the sequence follows a difference equation, Rn+2 = 6Rn+1 - Rn. It is noted that the last digit appears to be periodic with a cycle of 6, suggesting that R12345 has a units digit of 9. Further computational methods, including the Chinese remainder theorem and binomial theorem, are proposed for more complex calculations, but the focus remains on determining the last significant digit.
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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?
 
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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?

It is not difficult to find that the sequence is the solution of the difference equation...

$\displaystyle R_{n+2}= 6\ R_{n+1} - R_{n}$ (1)

... with the initial conditions $R_{0}=1$ and $R_{1}=3$. The first terms are...

$R_{0}=1\ ,\ R_{1}=3\ ,\ R_{2}= 17\ ,\ R_{3}=99\ ,\ R_{4}= 577\ ,\ R_{5}= 3363\ ,\ R_{6}= 21401\ ,\ ...$

... and that suggests that the last significant digit is periodic of period 6. Because is $12345 = 3\ \text {mod}\ 6$ the last significant digit of $R_{12345}$ should be 9...

Kind regards

$\chi$ $\sigma$
 
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.
 
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