$2x^2+3xy-4y^2=2$

hbghlyj

如何解 $2x^2+3xy-4y^2=2,(x,y\inZ)$

$(x,y)=(-6,-7)$
$(x,y)=(1,0)$

驗證：sagecell（在Pari文檔中qfbsolve）oeis

hejoseph

$2x^2+3xy-4y^2=2$ 等价于 $(4x+3y)^2-41y^2=16$，$t^2-41y^2=16$ 是广义 Pell 方程，其通解有三类
\begin{align*}
&t+y\sqrt{41}=\pm 4(2049+320\sqrt{41})^k\\
&t+y\sqrt{41}=\pm(45+7\sqrt{41})(2049+320\sqrt{41})^k\\
&t+y\sqrt{41}=\pm(365+57\sqrt{41})(2049+320\sqrt{41})^k\\
\end{align*}
其中 $k$ 为任意整数，简单讨论可得 $2x^2+3xy-4y^2=2$ 的通解为
\begin{align*}
&4x+3y+y\sqrt{41}=\pm 4(2049+320\sqrt{41})^k\\
&4x+3y+y\sqrt{41}=\pm(45+7\sqrt{41})(2049+320\sqrt{41})^k
\end{align*}
通解 $4x+3y+y\sqrt{41}=4(2049+320\sqrt{41})^k$ 取 $k=0$ 便得解 $(1,0)$。
通解 $4x+3y+y\sqrt{41}=-(45+7\sqrt{41})(2049+320\sqrt{41})^k$ 取 $k=0$ 便得解 $(-6,-7)$，取 $k=-1$ 便得除了上述解外绝对值最小的解 $(-134,57)$。

hbghlyj

sympygamma.com/input/?i=diophantine(2*x**2+3*x*y-4*y**2-2)

x	y
\[\frac{\left(-45 + 7 \sqrt{41}\right) \left(2049 - 320 \sqrt{41}\right)^{t}}{8} + \frac{\left(-45 - 7 \sqrt{41}\right) \left(320 \sqrt{41} + 2049\right)^{t}}{8} - \frac{3 \sqrt{41} \left(- \left(-45 + 7 \sqrt{41}\right) \left(2049 - 320 \sqrt{41}\right)^{t} + \left(-45 - 7 \sqrt{41}\right) \left(320 \sqrt{41} + 2049\right)^{t}\right)}{328}\]	\[\frac{\sqrt{41} \left(- \left(-45 + 7 \sqrt{41}\right) \left(2049 - 320 \sqrt{41}\right)^{t} + \left(-45 - 7 \sqrt{41}\right) \left(320 \sqrt{41} + 2049\right)^{t}\right)}{82}\]
\[\frac{\left(45 - 7 \sqrt{41}\right) \left(2049 - 320 \sqrt{41}\right)^{t}}{8} + \frac{\left(7 \sqrt{41} + 45\right) \left(320 \sqrt{41} + 2049\right)^{t}}{8} - \frac{3 \sqrt{41} \left(- \left(45 - 7 \sqrt{41}\right) \left(2049 - 320 \sqrt{41}\right)^{t} + \left(7 \sqrt{41} + 45\right) \left(320 \sqrt{41} + 2049\right)^{t}\right)}{328}\]	\[\frac{\sqrt{41} \left(- \left(45 - 7 \sqrt{41}\right) \left(2049 - 320 \sqrt{41}\right)^{t} + \left(7 \sqrt{41} + 45\right) \left(320 \sqrt{41} + 2049\right)^{t}\right)}{82}\]
\[\frac{\left(-8196 + 1280 \sqrt{41}\right) \left(2049 - 320 \sqrt{41}\right)^{t}}{8} + \frac{\left(-8196 - 1280 \sqrt{41}\right) \left(320 \sqrt{41} + 2049\right)^{t}}{8} - \frac{3 \sqrt{41} \left(- \left(-8196 + 1280 \sqrt{41}\right) \left(2049 - 320 \sqrt{41}\right)^{t} + \left(-8196 - 1280 \sqrt{41}\right) \left(320 \sqrt{41} + 2049\right)^{t}\right)}{328}\]	\[\frac{\sqrt{41} \left(- \left(-8196 + 1280 \sqrt{41}\right) \left(2049 - 320 \sqrt{41}\right)^{t} + \left(-8196 - 1280 \sqrt{41}\right) \left(320 \sqrt{41} + 2049\right)^{t}\right)}{82}\]
\[\frac{\left(2049 - 320 \sqrt{41}\right)^{t} \left(8196 - 1280 \sqrt{41}\right)}{8} + \frac{\left(320 \sqrt{41} + 2049\right)^{t} \left(1280 \sqrt{41} + 8196\right)}{8} - \frac{3 \sqrt{41} \left(- \left(2049 - 320 \sqrt{41}\right)^{t} \left(8196 - 1280 \sqrt{41}\right) + \left(320 \sqrt{41} + 2049\right)^{t} \left(1280 \sqrt{41} + 8196\right)\right)}{328}\]	\[\frac{\sqrt{41} \left(- \left(2049 - 320 \sqrt{41}\right)^{t} \left(8196 - 1280 \sqrt{41}\right) + \left(320 \sqrt{41} + 2049\right)^{t} \left(1280 \sqrt{41} + 8196\right)\right)}{82}\]

hbghlyj

hejoseph 发表于 2024-2-29 00:55
其通解有三类

这个是怎么算出来的呢？

hbghlyj

alpertron.com.ar/QUAD.HTM

2 ⁢x² + 3 ⁢x⁢y - 4 ⁢y² - 2 ⁢ = 0

The discriminant is D = b² − 4⁢a⁢c = 41

2 X² + 3 X⁢Y − 4 ⁢Y² = 2 (1)

The algorithm requires that the coefficient of X² and the right hand side are coprime. This does not happen, so we have to find a value of m such that applying one of the unimodular transformations

• X = m⁢U + (m−1)⁢V, Y = U + V
• X = U + V, Y = (m−1)⁢U + m⁢V

the coefficient of U² and the right hand side are coprime. This coefficient equals 2 m² + 3 m − 4 in the first case and -4 (m − 1)² + 3 (m − 1) + 2 in the second case.

We will use the first unimodular transformation with m = 1: X = U, Y = U + V (2)

Using (2), the equation (1) converts to:

U² − 5 U⁢V − 4 ⁢V² = 2 (3)

We will have to solve several quadratic modular equations. To do this we have to factor the modulus and find the solution modulo the powers of the prime factors. Then we combine them by using the Chinese Remainder Theorem.

The different moduli are divisors of the right hand side, so we only have to factor it once.

2 = 2

Searching for solutions U and V coprime.

We have to solve: T² − 5 T − 4 ≡ 0 (mod 2 = 2)

Solutions modulo 2: 0 and 1

1. T = 0

The transformation U = - 2 ⁢k (4) converts U² − 5 U⁢V − 4 ⁢V² = 2 to P⁢V² + Q⁢Vk + R⁢ k² = 1 (5)

where: P = (a⁢T² + b⁢T + c) / n = -2, Q = −(2⁢a⁢T + b) = 5, R = a⁢n = 2

The continued fraction expansion of (Q + D) / (2R) = (5 + 41) / 4 is:

2+ // 1, 5, 1, 2, 2// (6)
Solution of (5) found using the convergent V / (−k) = 1 / 3 of (6)

U = 6, V = 1

From (2):

X = 6, Y = 7

x = 6
y = 7

U = -6, V = -1

From (2):

X = -6, Y = -7

x = -6
y = -7

The continued fraction expansion of (−Q + D) / (−2R) = (-5 + 41) / (-4) is:

-1+ // 1, 1, 1, 5, 1, 2, 2// (7)

x = 6
y = 7

x = -6
y = -7

2. T = 1

The transformation U =  V - 2 ⁢k (8) converts U² − 5 U⁢V − 4 ⁢V² = 2 to P⁢V² + Q⁢Vk + R⁢ k² = 1 (9)

where: P = (a⁢T² + b⁢T + c) / n = -4, Q = −(2⁢a⁢T + b) = 3, R = a⁢n = 2

The continued fraction expansion of (Q + D) / (2R) = (3 + 41) / 4 is:

2+ // 2, 1, 5, 1, 2// (10)

The continued fraction expansion of (−Q + D) / (−2R) = (-3 + 41) / (-4) is:

-1+ // 6, 1, 2, 2, 1, 5// (11)
Solution of (9) found using the convergent V / (−k) = 1 / -1 of (11)

U = -1, V = 1

From (2):

X = -1, Y = 0

x = -1
y = 0

U = 1, V = -1

From (2):

X = 1, Y = 0

x = 1
y = 0

x = -1
y = 0

x = 1
y = 0

Recursive solutions:

x_n+1 = 1089 ⁢x_n + 2560 ⁢y_n
y_n+1 = 1280 ⁢x_n + 3009 ⁢y_n

and also:

x_n+1 = 3009 ⁢x_n - 2560 ⁢y_n
y_n+1 = - 1280 ⁢x_n + 1089 ⁢y_n

hejoseph

hbghlyj 发表于 2024-10-3 16:38
这个是怎么算出来的呢？

找《谈谈不定方程》这本书，里面有具体解法。