solving $x^{2026}+\frac{1}{x^{2026}}$ given $x+\frac{1}{x}=1$

Question

if $x+\frac{1}{x}=1$, solve $x^{2026}+\frac{1}{x^{2026}}=$?​

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Answer

From $x+\frac{1}{x}=1$, multiply through by $x$ (note $x\ne 0$):

Method 1 — recurrence (structure):​ Define $a_n=x^n+\frac{1}{x^n}$. Then

Starting with $a_0=2,a_1=1$:

The sequence repeats with period 6. Since $2026\equiv 4(\mathrm{mod}6)$:

Method 2 — roots of unity (cross-check):​ $x^2-x+1=0$ gives $x=\frac{1\pm i\sqrt{3}}{2}=e^{\pm i\pi /3}$, so $x^6=1$ and $x^3=-1$. Then

more directly: $x^4+x^{-4}=x\cdot x^3+x^{-3}\cdot x=-x-x=-(x+x^{-1})=-1$. ✓

Both methods agree.

• Key concepts
  • $x+\frac{1}{x}=1$ is equivalent to $x^2-x+1=0$, whose roots satisfy $x^3=-1$ and $x^6=1$ — a 6th root of unity.
  • The sequence $a_n=x^n+x^{-n}$ satisfies $a_n=a_{n-1}-a_{n-2}$, making the pattern periodic with period 6.
• Confusion points
  • Computing $x^{2026}$ by "2026 mod 6 = 4, so use $x^4$" — correct, but only because $x^6=1$; a common trap is using $x^3=1$ (which holds for ordinary cube roots of unity) instead of $x^3=-1$ here.
  • Forgetting that the recurrence needs two seeds: $a_0=2$ and $a_1=1$; using only $a_1$ gives the wrong phase of the cycle.