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I am trying to find the five smallest pairs of positive integers $p,q$ that satisfy the Pell's equation $p^2-321q^2 = 1$.

One obvious trivial solution is $p=1$ and $q=0$ , but this does not count. I am trying to find the five smallest pairs of positive integers $p,q$ that satisfy this equation. I would appreciate your help .

kimchi lover
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Derak
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    The fundamental solution is $(215/12)$. To find it, look at the convergents corresponding to the simple continued fraction of $\sqrt{321}$ – Peter Jan 03 '20 at 20:21

4 Answers4

4

Method suitable for hand calculations.

As in the other answer, the larger solutions come from powers of the matrix $$ A = \left( \begin{array}{cc} 215 & 3852 \\ 12 & 215 \\ \end{array} \right) $$ The entries in $A^2, A^3, A^4, A^5$ become quite large; still, the main diagonals have two equal numbers, and the determinants remain $1.$

The letter $A$ stands for Automorphism, this matrix (notice that the determinant is $1\;$) is the generator of the (oriented) automorphism group of the quadratic form $x^2 - 321 y^2.$

Method described by Prof. Lubin at Continued fraction of $\sqrt{67} - 4$

$$ \sqrt { 321} = 17 + \frac{ \sqrt {321} - 17 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {321} - 17 } = \frac{ \sqrt {321} + 17 }{32 } = 1 + \frac{ \sqrt {321} - 15 }{32 } $$ $$ \frac{ 32 }{ \sqrt {321} - 15 } = \frac{ \sqrt {321} + 15 }{3 } = 10 + \frac{ \sqrt {321} - 15 }{3 } $$ $$ \frac{ 3 }{ \sqrt {321} - 15 } = \frac{ \sqrt {321} + 15 }{32 } = 1 + \frac{ \sqrt {321} - 17 }{32 } $$ $$ \frac{ 32 }{ \sqrt {321} - 17 } = \frac{ \sqrt {321} + 17 }{1 } = 34 + \frac{ \sqrt {321} - 17 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccc} & & 17 & & 1 & & 10 & & 1 & & 34 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 17 }{ 1 } & & \frac{ 18 }{ 1 } & & \frac{ 197 }{ 11 } & & \frac{ 215 }{ 12 } \\ \\ & 1 & & -32 & & 3 & & -32 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 321 \cdot 0^2 = 1 & \mbox{digit} & 17 \\ \frac{ 17 }{ 1 } & 17^2 - 321 \cdot 1^2 = -32 & \mbox{digit} & 1 \\ \frac{ 18 }{ 1 } & 18^2 - 321 \cdot 1^2 = 3 & \mbox{digit} & 10 \\ \frac{ 197 }{ 11 } & 197^2 - 321 \cdot 11^2 = -32 & \mbox{digit} & 1 \\ \frac{ 215 }{ 12 } & 215^2 - 321 \cdot 12^2 = 1 & \mbox{digit} & 34 \\ \end{array} $$

Will Jagy
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All solutions of equation $p^2−321q^2=1$ derives from modulo polynomial $p+qx\equiv(12x-215)^j\pmod{x^2-321}$ where $j\in\mathbb Z^+$

(215, 12)
(92449, 5160)
(39752855, 2218788)
(17093635201, 954073680)
(7350223383575, 410249463612)
(3160578961302049, 176406315279480)
(1359041603136497495, 75854305320712788)
(584384728769732620801, 32617174881591219360)
(251284074329381890446935, 14025309344778903612012)
(108051567576905443159561249, 6030850401080046961945800)
Dmitry Ezhov
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Extension of Will Jagy's answer:

given the fundamental solution $P,Q$ then all solutions are given by $P_n, Q_n$, where $n \in \mathbb{Z^+}$ and $(P + Q\sqrt{321})^n = P_n + Q_n\sqrt{321}$

user2661923
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If you want an elementary method, you can write:

$p^2-D q^2=1$

$(p-1)(p+1)=D q^2$

If D can be factorized, then p-1 or p+1 must have common divisors with one or two factors of D. For example :

$p^2-321q^2=1$

⇒ $(p-1)(p+1)=3\times 107 q^2$

321 is odd so one of p or q must be even. Let q be even then we can write:

$(p-1)(p+1)=3\times 107\times 2^2 q_1^2$

This indicates that LHS must also have two even factors. so we must have:

$(p_1-1)(p_1+1)=3\times (2 \times 107)2\times q^2 $

Suppose $p_1-1=214$ ⇒ $p_1=215$ ⇒ $q=12$

Where $p ≡0\ mod(p_1)$ ⇒ $ p-1≡0 \ mod (107)$

Also $q ≡ 0 \mod 12$

You see this in solutions given by Dmitry.

sirous
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