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  "body": "\n## Proof\n\nLet $\\mathbb{Z}[\\sqrt{d}] = \\set{a + b\\sqrt{d} : a, b \\in \\Z}$.\n\n1. Since $d$ is not a perfect square, $\\sqrt{d}$ is irrational by [[the irrationality of square roots of non-squares|0N10YRNNWV]]. So $a + b\\sqrt{d} = a' + b'\\sqrt{d}$ with integers $a, b, a', b'$ implies $a = a'$ and $b = b'$.\n2. For $\\gamma = a + b\\sqrt{d}$ in $\\mathbb{Z}[\\sqrt{d}]$, put $\\bar\\gamma = a - b\\sqrt{d}$ and $N(\\gamma) = \\gamma \\bar\\gamma = a^2 - d b^2$. By step 1 these are well defined.\n3. The set $\\mathbb{Z}[\\sqrt{d}]$ is closed under products. For $\\gamma = a + b\\sqrt{d}$ and $\\delta = c + e\\sqrt{d}$, $\\gamma\\delta = (ac + dbe) + (ae + bc)\\sqrt{d}$ and $\\bar\\gamma \\bar\\delta = (ac + dbe) - (ae + bc)\\sqrt{d}$. So $\\overline{\\gamma\\delta} = \\bar\\gamma \\bar\\delta$ and $N(\\gamma\\delta) = N(\\gamma) N(\\delta)$.\n4. So a pair $(a, b)$ of integers solves [[Pell's equation|0AF3DANJ76]] exactly when $N(a + b\\sqrt{d}) = 1$, and products of such numbers again have norm $1$.\n5. Lemma 1: if $N(\\eta) = 1$ and $\\eta = a + b\\sqrt{d} \u003e 1$, then $a \u003e 1$ and $b \\ge 1$. Indeed $\\bar\\eta = 1/\\eta \\in (0, 1)$. So $a = (\\eta + \\bar\\eta)/2 \u003e 0$ and $b\\sqrt{d} = (\\eta - \\bar\\eta)/2 \u003e 0$. Then $b \\ge 1$ and $a^2 = 1 + db^2 \u003e 1$, so $a \u003e 1$.\n6. Lemma 2: let $(a, b)$ and $(c, e)$ be solutions with $a, b, c, e \\ge 1$ and $a \u003c c$. Then $b \u003c e$, because $d b^2 = a^2 - 1 \u003c c^2 - 1 = d e^2$. Hence $a + b\\sqrt{d} \u003c c + e\\sqrt{d}$.\n7. Let $\\varepsilon = x_1 + y_1\\sqrt{d}$. By the definition of the [[fundamental solution|0KCYJSB9FW]], $x_1 \u003e 1$ and $y_1 \u003e 0$, so $\\varepsilon \u003e 1$, and $N(\\varepsilon) = 1$.\n8. By step 3, $\\bar\\varepsilon = x_1 - y_1\\sqrt{d}$ lies in $\\mathbb{Z}[\\sqrt{d}]$ and $\\varepsilon \\bar\\varepsilon = 1$. So $\\varepsilon^{-n} = \\bar\\varepsilon^{\\,n}$ lies in $\\mathbb{Z}[\\sqrt{d}]$ and has norm $1$ for each $n \\ge 0$.\n9. Let $\\xi = x + y\\sqrt{d}$. Since $x \\ge 1$ and $y \\ge 0$, we have $\\xi \\ge 1$.\n10. Since $\\varepsilon \u003e 1$, the powers $\\varepsilon^0 = 1 \u003c \\varepsilon \u003c \\varepsilon^2 \u003c \\cdots$ increase. By induction on $n$, $\\varepsilon^n \\ge 1 + n(\\varepsilon - 1)$. By the [[Archimedean property|0A3GWZ86AY]], the powers $\\varepsilon^n$ are unbounded.\n11. So there is a largest natural number $n$ with $\\varepsilon^n \\le \\xi$, and for it $\\varepsilon^n \\le \\xi \u003c \\varepsilon^{n+1}$.\n12. Put $\\zeta = \\xi \\varepsilon^{-n} = \\xi \\bar\\varepsilon^{\\,n}$. By steps 3 and 8, $\\zeta = a + b\\sqrt{d}$ with integers $a$, $b$, and $N(\\zeta) = 1$.\n13. Dividing step 11 by $\\varepsilon^n \u003e 0$ gives $1 \\le \\zeta \u003c \\varepsilon$.\n14. Suppose $\\zeta \u003e 1$. By Lemma 1, $a \u003e 1$ and $b \\ge 1$, so $(a, b)$ is a solution with $a \u003e 1$ and $b \u003e 0$. By the minimality of $x_1$, $a \\ge x_1$.\n15. If $a = x_1$, then $d b^2 = a^2 - 1 = d y_1^2$, so $b = y_1$ and $\\zeta = \\varepsilon$. If $a \u003e x_1$, then $\\zeta \u003e \\varepsilon$ by Lemma 2. Both contradict step 13.\n16. So $\\zeta = 1$, that is, $x + y\\sqrt{d} = \\varepsilon^n = (x_1 + y_1\\sqrt{d})^n$.\n17. Uniqueness: if also $\\xi = \\varepsilon^m$ with $m \\ge 0$, then $\\varepsilon^m = \\varepsilon^n$. Since the powers of $\\varepsilon$ strictly increase, $m = n$.\n\n## Idea\n\nSolutions of Pell's equation are the numbers $a + b\\sqrt{d}$ of norm $1$, and these are closed under products and inverses.\nDivide a given solution by the largest power of $\\varepsilon = x_1 + y_1\\sqrt{d}$ below it.\nThe quotient is a solution between $1$ and $\\varepsilon$, and no solution other than $1$ lies there, because $\\varepsilon$ is the smallest solution above $1$.\n",
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