Algebra problem (x+(1/x) problem)

[terryds]

Homework Statement

If ##a+\frac{1}{a}=\sqrt{3}##, then ##a^{2016} + (\frac{1}{a})^{2016} ## equals

A. 0
B. 1
C. √2
D. √3
E. 2

Homework Equations

The Attempt at a Solution

[/B]
I find no real solution for a.
What I can do is just find a^2 + (1/a)^2 which equals 1, a^4 + (1/a)^4 = -1, a^8 + (1/a)^8 = -1, and so on...
Since 2016 is not a square number, I don't know how to determine the value

Is there any theory about problem like this?
Please help

 

[Samy_A]

Homework Statement

If ##a+\frac{1}{a}=\sqrt{3}##, then ##a^{2016} + (\frac{1}{a})^{2016} ## equals

A. 0
B. 1
C. √2
D. √3
E. 2

Homework Equations

The Attempt at a Solution

[/B]
I find no real solution for a.
What I can do is just find a^2 + (1/a)^2 which equals 1, a^4 + (1/a)^4 = -1, a^8 + (1/a)^8 = -1, and so on...
Since 2016 is not a square number, I don't know how to determine the value

Is there any theory about problem like this?
Please help

Start from ##a²+1/{a²}=1##.
Multiply by ##a²## to get rid of the denominator, you get ##a^4 +1 =a²##.
Now notice that 2016 is a multiple of 6. So multiply ##a^4 +1 =a²## again by ##a²## to get ##a^6## in the equation.

 

[ehild]

The Attempt at a Solution

[/B]
I find no real solution for a.

Find the complex solution. Write it in trigonometric of exponential form, which makes it easy to rise to power 2016. (2016=2⁵*3² *7)

 

[terryds]

Start from ##a²+1/{a²}=1##.
Multiply by ##a²## to get rid of the denominator, you get ##a^4 +1 =a²##.
Now notice that 2016 is a multiple of 6. So multiply ##a^4 +1 =a²## again by ##a²## to get ##a^6## in the equation.

a^6 + a^2 = a^4
a^6 = a^4 - a^2 = (a^2 - 1) - a^2 = -1
a^12 = 1
a^24 = 1
...
a^2016 = 1

So, 1/a^2016 = 1
a^2016 + (1/a)^2016 = 1 + 1 = 2
Thank you

 

[terryds]

Find the complex solution. Write it in trigonometric of exponential form, which makes it easy to rise to power 2016. (2016=2⁵*3² *7)

I'm interested to try this approach..

The solution for a is ##\frac{\sqrt{3}\pm i}{2}##
Should I use the plus or the minus as the root?
If it's plus, ##e^{i\frac{1}{6}\pi}##
If it's minus, ##e^{i\frac{11}{6}\pi}##

And, then what should I do??

 

[ehild]

I'm interested to try this approach..

The solution for a is ##\frac{\sqrt{3}\pm i}{2}##
Should I use the plus or the minus as the root?
If it's plus, ##e^{i\frac{1}{6}\pi}##
If it's minus, ##e^{i\frac{11}{6}\pi}##

And, then what should I do??

What are their 12th power?

 

[terryds]

What are their 12th power?

The 'plus' root^12 = ##e^{i2\pi}= 1##
The 'minus' root^12 = ##e^{i22\pi} = 1##

Okay, since 2016 can be divided by 12, a^2016 = 1
And 1/a^2016 = 1
So, a^2016 + 1/a^2016 = 1+1 = 2

Thanks a lot!

 

[ehild]

The 'plus' root^12 = ##e^{i2\pi}= 1##
The 'minus' root^12 = ##e^{i22\pi} = 1##

Okay, since 2016 can be divided by 12, a^2016 = 1
And 1/a^2016 = 1
So, a^2016 + 1/a^2016 = 1+1 = 2

Thanks a lot!

The 'minus' root^12 = ##e^{-i2\pi} = 1##