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  "body": "\n## Proof\n\n1. Since $d$ is not a perfect square, $\\sqrt{d}$ is irrational by [[the irrationality of square roots of non-squares|0N10YRNNWV]].\n2. Claim: there are infinitely many pairs of integers $(p, q)$ with $q \\ge 1$ and $\\abs{p - q\\sqrt{d}} \u003c 1/q$.\n3. For a positive integer $N$, [[Dirichlet's approximation theorem|04JAY9DGB7]] gives integers $p$, $q$ with $1 \\le q \\le N$ and $\\abs{q\\sqrt{d} - p} \\le \\frac{1}{N+1} \u003c \\frac{1}{N} \\le \\frac{1}{q}$. So at least one such pair exists.\n4. Suppose only finitely many pairs had this property. By step 1, $p - q\\sqrt{d} \\ne 0$ for each of them. Let $\\mu \u003e 0$ be the least value of $\\abs{p - q\\sqrt{d}}$ over them.\n5. Choose $N$ with $\\frac{1}{N+1} \u003c \\mu$. Step 3 gives a pair with the property and $\\abs{p - q\\sqrt{d}} \u003c \\mu$. It is not in the finite list, a contradiction. This proves the claim.\n6. For each pair of step 2, $\\abs{p + q\\sqrt{d}} \\le \\abs{p - q\\sqrt{d}} + 2q\\sqrt{d} \u003c \\frac{1}{q} + 2q\\sqrt{d}$.\n7. So $\\abs{p^2 - d q^2} = \\abs{p - q\\sqrt{d}} \\cdot \\abs{p + q\\sqrt{d}} \u003c \\frac{1}{q}\\left(\\frac{1}{q} + 2q\\sqrt{d}\\right) \\le 1 + 2\\sqrt{d}$.\n8. Also $p^2 - d q^2 \\ne 0$, since otherwise $\\sqrt{d} = \\abs{p}/q$ would be rational.\n9. So $p^2 - dq^2$ takes values in the finite set of nonzero integers $k$ with $\\abs{k} \u003c 1 + 2\\sqrt{d}$. By the [[infinite pigeonhole principle|0FS3RHF6KE]], some such $k$ equals $p^2 - dq^2$ for infinitely many pairs of step 2.\n10. Each such pair gives a pair of residues $(p \\bmod \\abs{k}, q \\bmod \\abs{k})$, and there are only $k^2$ of these. By the [[pigeonhole principle|0HGVCWQGY3]], there are two different pairs $(p_1, q_1) \\ne (p_2, q_2)$ among them with $p_1 \\equiv p_2$ and $q_1 \\equiv q_2 \\pmod{\\abs{k}}$.\n11. Put $X = p_1 p_2 - d q_1 q_2$ and $Y = q_1 p_2 - p_1 q_2$. Then $(p_1 + q_1\\sqrt{d})(p_2 - q_2\\sqrt{d}) = X + Y\\sqrt{d}$.\n12. Modulo $\\abs{k}$, step 10 gives $X \\equiv p_1^2 - d q_1^2 = k \\equiv 0$ and $Y \\equiv q_1 p_1 - p_1 q_1 = 0$. So $X = kx$ and $Y = ky$ for integers $x$ and $y$.\n13. Expanding both sides, as in the [[Brahmagupta–Fibonacci identity|0K2DMACEZ4]], gives $X^2 - dY^2 = (p_1^2 - d q_1^2)(p_2^2 - d q_2^2) = k^2$.\n14. By steps 12 and 13, $k^2 (x^2 - d y^2) = k^2$. Since $k \\ne 0$, $x^2 - d y^2 = 1$.\n15. Suppose $y = 0$. Then $Y = 0$, so $p_2/q_2 = p_1/q_1$. Put $t = q_2/q_1 \u003e 0$; then $p_2 = t p_1$ and $q_2 = t q_1$.\n16. Then $k = p_2^2 - d q_2^2 = t^2 (p_1^2 - d q_1^2) = t^2 k$, so $t^2 = 1$ and $t = 1$. This gives $(p_2, q_2) = (p_1, q_1)$, contrary to step 10. So $y \\ne 0$.\n17. So [[Pell's equation|0AF3DANJ76]] has a solution $(x, y)$ with $y \\ne 0$. Replacing $x$ by $\\abs{x}$ and $y$ by $\\abs{y}$, we may take $x \\ge 1$ and $y \\ge 1$.\n18. For infinitely many solutions, put $u = x + y\\sqrt{d}$. Then $u \\ge 1 + \\sqrt{d} \u003e 1$.\n19. Define integers $x_n$, $y_n$ for $n \\ge 0$ by $x_0 = 1$, $y_0 = 0$, $x_{n+1} = x x_n + d y y_n$ and $y_{n+1} = y x_n + x y_n$.\n20. By induction on $n$, $x_n + y_n \\sqrt{d} = u^n$ and $x_n - y_n\\sqrt{d} = (x - y\\sqrt{d})^n$. Both follow by multiplying out $(x \\pm y\\sqrt{d})(x_n \\pm y_n\\sqrt{d})$.\n21. Multiplying the two formulas of step 20 gives $x_n^2 - d y_n^2 = (x^2 - d y^2)^n = 1$. So each $(x_n, y_n)$ is a solution.\n22. Since $u \u003e 1$, the numbers $u^n$ are pairwise different. So the pairs $(x_n, y_n)$ are pairwise different, and Pell's equation has infinitely many integer solutions.\n\n## Idea\n\nDirichlet's theorem gives infinitely many fractions $p/q$ near $\\sqrt{d}$, and for them $p^2 - dq^2$ stays bounded.\nSo one value $k$ repeats infinitely often, and two of these pairs agree modulo $k$.\nDividing one by the other, in the form $(p_1 + q_1\\sqrt{d})(p_2 - q_2\\sqrt{d})/k$, gives a solution of norm $1$.\nPowers of one solution then give infinitely many.\n",
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