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  "body": "\n## Proof\n\nWe argue by [[induction|0ASJVEFA9J]] on $n$, for two consecutive terms at a time.\n\n1. Let $\\varphi = \\frac{1 + \\sqrt{5}}{2}$ and $\\psi = \\frac{1 - \\sqrt{5}}{2}$. For a natural number $n$, let $G_n = \\frac{\\varphi^n - \\psi^n}{\\sqrt{5}}$.\n2. By the description of the [[golden ratio|0SC8T7548D]], $\\varphi$ and $\\psi$ are the roots of $x^2 - x - 1$. So $\\varphi^2 = \\varphi + 1$ and $\\psi^2 = \\psi + 1$.\n3. Directly, $\\psi^2 = \\frac{1 - 2\\sqrt{5} + 5}{4} = \\frac{3 - \\sqrt{5}}{2} = \\psi + 1$, and the same computation with $+\\sqrt{5}$ checks $\\varphi^2 = \\varphi + 1$.\n4. We have $G_0 = \\frac{1 - 1}{\\sqrt{5}} = 0$ and $G_1 = \\frac{\\varphi - \\psi}{\\sqrt{5}} = \\frac{\\sqrt{5}}{\\sqrt{5}} = 1$.\n5. For every natural number $n$, step 2 gives $\\varphi^{n+2} = \\varphi^{n+1} + \\varphi^n$ and $\\psi^{n+2} = \\psi^{n+1} + \\psi^n$. Subtracting and dividing by $\\sqrt{5}$ gives $G_{n+2} = G_{n+1} + G_n$.\n6. Let $P(n)$ be the claim that $G_n = F_n$ and $G_{n+1} = F_{n+1}$, where $F_m$ are the [[Fibonacci numbers|0GA9K57A39]].\n7. $P(0)$ holds by step 4, since $F_0 = 0$ and $F_1 = 1$.\n8. Assume $P(n)$. Then $G_{n+1} = F_{n+1}$, and by step 5, $G_{n+2} = G_{n+1} + G_n = F_{n+1} + F_n = F_{n+2}$. So $P(n + 1)$ holds.\n9. By induction, $P(n)$ holds for every natural number $n$. In particular $F_n = \\frac{\\varphi^n - \\psi^n}{\\sqrt{5}}$ for every $n$.\n\n## Idea\n\nSince $\\varphi$ and $\\psi$ satisfy $x^2 = x + 1$, their powers obey the Fibonacci rule. So does every combination $a \\varphi^n + b \\psi^n$. The choice $a = -b = \\frac{1}{\\sqrt{5}}$ matches the starting values $0$ and $1$, so it gives the Fibonacci numbers.\n",
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