Find $z^5 +\frac{1}{z^5}$ given that $z > 0$ and $z^2 + \frac{1}{z^2}=14$ I have been struggling with this problem and I think I am nearly there. Here is my working so far:



Any hints?
As $z&gt;0, z+\dfrac1z=4$

There is a typo in your $$\left(z+\dfrac1z\right)^3$$

The right hand side should be $$z^3+\dfrac1{z^3}+3\left(z+\dfrac1z\right)$$

which will give us the value of $z^3+\dfrac1{z^3}$

Finally,

$$\left(z^2+\dfrac1{z^2}\right)\left(z^3+\dfrac1{z^3}\right)=z^5+\dfrac1{z^5}+\left(z+\dfrac1{z}\right)$$

Replace the values of $z+\dfrac1{z},z^2+\dfrac1{z^2}, z^3+\dfrac1{z^3}$

See also : Proving that $\frac{\phi^{400}+1}{\phi^{200}}$ is an integer.
As $z&gt;0,z+\dfrac1z=4$

Consequently $z,\dfrac1z$ are the roots of $$t^2-4t+1=0$$

$$t^{n+2}=4t^{n+1}-t^n$$

If $z^m+\dfrac1{z^m}=S_m$

$$S_{n+2}=4S_{n+1}-S_n$$ with $S_0=1+1$ and $S_1=4,S_2=14$

Set $n=1,2,3$