If x^2+x+1=0, then the value of [(x+1x)^4+(x^2+1x^2)^4+(x^3+1x^3)^4++(x^25+1x^25)^4] is:

Question

If $x^2+x+1=0$, then the value of
$$\left(x+\frac1x\right)^4+\left(x^2+\frac1{x^2}\right)^4+\left(x^3+\frac1{x^3}\right)^4+\cdots+\left(x^{25}+\frac1{x^{25}}\right)^4$$
is:

Accepted Answer

The roots of $x^2+x+1=0$ are $\omega,\omega^2$ (non-real cube roots of unity). Take $x=\omega$. For any integer $k$,
$$x^k+\frac1{x^k}=\omega^k+\omega^{-k}=\omega^k+\omega^{2k}.$$

- If $3\mid k$: $\omega^k=1$, so this equals $1+1=2$, and the fourth power is $2^4=16$.
- If $3\nmid k$: $\omega^k,\omega^{2k}$ are the two non-real cube roots, so $\omega^k+\omega^{2k}=-1$ (since $1+\omega^k+\omega^{2k}=0$), and the fourth power is $(-1)^4=1$.

Among $k=1,\dots,25$, the multiples of 3 are $3,6,\dots,24$ --- that's 8 values; the remaining $25-8=17$ values are not multiples of 3.

$$\text{Sum}=17(1)+8(16)=17+128=145.$$

Solutions · 1