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  "body": "\n## Proof\n\nWe argue by [[induction from the base case $n = 1$|0GF6AJEBZ0]].\n\n1. Let $F_m$ be the [[Fibonacci numbers|0GA9K57A39]], with $F_0 = 0$, $F_1 = 1$ and $F_{m+2} = F_{m+1} + F_m$. For $n \\ge 1$, let $P(n)$ be the claim $F_{n-1} F_{n+1} - F_n^2 = (-1)^n$.\n2. Base case. $F_2 = F_1 + F_0 = 1$, so $F_0 F_2 - F_1^2 = 0 - 1 = -1 = (-1)^1$. Thus $P(1)$ holds.\n3. Induction step. Let $n \\ge 1$ and assume $P(n)$.\n4. By the recurrence, $F_{n+2} = F_{n+1} + F_n$. So $F_n F_{n+2} - F_{n+1}^2 = F_n F_{n+1} + F_n^2 - F_{n+1}^2$.\n5. Rearranging, this equals $F_n^2 - F_{n+1}(F_{n+1} - F_n)$.\n6. Since $n \\ge 1$, the recurrence gives $F_{n+1} = F_n + F_{n-1}$, so $F_{n+1} - F_n = F_{n-1}$.\n7. Hence $F_n F_{n+2} - F_{n+1}^2 = F_n^2 - F_{n-1} F_{n+1} = -(F_{n-1} F_{n+1} - F_n^2)$.\n8. By $P(n)$, this equals $-(-1)^n = (-1)^{n+1}$. This is $P(n + 1)$.\n9. By induction, $F_{n-1} F_{n+1} - F_n^2 = (-1)^n$ for every positive integer $n$.\n\n## Idea\n\nReplace $F_{n+2}$ by $F_{n+1} + F_n$ in the next instance of the identity. After one more use of the recurrence, the expression becomes the previous one with the opposite sign. So the sign alternates, starting from $-1$ at $n = 1$.\n",
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