solve-x-2-7y-2-1-in-Z-

Question Number 739 by malwaan last updated on 08/Mar/15

s o l v e x^{2} - 7 y^{2} = 1 i n Z

Commented by 123456 last updated on 06/Mar/15

S = {(x, y) \in Z^{2} ∣ x^{2} - 7 y^{2} = 1}

Commented by 123456 last updated on 06/Mar/15

(x, y) = (- 1, 0)

(x, y) = (1, 0)

(x, y) = (8, 3)

Commented by 123456 last updated on 06/Mar/15

i f (x_{1}, y_{1}) i s a s o l u t i o n, t h e n (x_{1}, - y_{1})

i s a s o l u t i o n

x_{1}^{2} - 7 {(\pm y_{1})}^{2} = x_{1}^{2} - 7 y_{1}^{2}

Commented by 123456 last updated on 06/Mar/15

(u^{2} - 7 v^{2}) (x^{2} - 7 y^{2})

= u^{2} x^{2} - 7 u^{2} y^{2} - 7 v^{2} x^{2} + 49 v^{2} y^{2}

= {(u x)}^{2} + {(7 v y)}^{2} - 7 [{(u y)}^{2} + {(v x)}^{2}]

= {(u x)}^{2} + {(7 v y)}^{2} - 7 [{(u y)}^{2} + {(v x)}^{2}] + 2 (u x) (7 v y) - 2 (u x) (v y)

= {(u x)}^{2} \pm 2 (u x) (7 v y) + {(7 v y)}^{2} - 7 [{(u y)}^{2} \pm 2 (u y) (v x) + {(v x)}^{2}]

= {(u x \pm 7 v y)}^{2} - 7 {(u y \pm v x)}^{2}

f : Z^{4} \to Z^{2}

f [(x_{1}, y_{1}), (x_{2}, y_{2})] = (x_{1} x_{2} + 7 y_{1} y_{2}, x_{1} y_{2} + x_{2} y_{1})

Answered by prakash jain last updated on 06/Mar/15

7 y^{2} = x^{2} - 1 = (x + 1) (x - 1)

7 divides (x + 1) (x - 1) \Rightarrow x = 7 k + 1 or 7 k - 1

Case I : x = 7 k + 1

7 y^{2} = 7 k (7 k + 2) \Rightarrow y^{2} = k (7 k + 2)

k (7 k + 2) is perfect square so k is a factor of

7 k + 2 .

\frac{7 k + 2}{k} = 7 + \frac{2}{k}

Only possible values for k are 0, \pm 1, \pm 2

k = 0, k (7 k + 2) = 0 = 0^{2} \Rightarrow y = 0, x = 7 k + 1 = 1

k = 1, k (7 k + 2) = 9 = 3^{2} \Rightarrow y = \pm 3, x = 7 k + 1 = 8

k = - 1, k (7 k + 2) = 5 (not a perfect square)

k = 2, k (7 k + 2) = 32 (not a perfect square)

k = - 2, k (7 k + 2) = 24 (not a perfect square)

Case II : x = 7 k - 1

y^{2} = k (7 k - 2)

As above k = 0, \pm 1, \pm 2

k = 0, k (7 k - 2) = 0 = 0^{2} \Rightarrow y = 0, x = 7 k - 1 = - 1

k = 1, k (7 k - 2) = 5 (not a perfect square)

k = - 1, k (7 k - 2) = 9 = 3^{2} \Rightarrow y = \pm 3, x = 7 k - 1 = - 8

k = 2, k (7 k - 2) = 24 (not a perfect square)

k = - 2, k (7 k - 2) = 32 (not a perfect square)

So the only solution x^{2} - 7 y^{2} = 1 are

x = - 1, y = 0

x = 1, y = 0

x = 8, y = \pm 3

x = - 8, y = \pm 3