Question-113510

Question Number 113510 by Lekhraj last updated on 13/Sep/20

Answered by MJS_new last updated on 13/Sep/20

x^{3} - y^{3} + 9 x y + 127 = 0

x, y, p \in Z

let y = x + p

(9 - 3 p) x^{2} + (9 p - 3 p^{2}) x - p^{3} + 127 = 0

x^{2} + p x + \frac{p^{3} - 127}{3 (p - 3)} = 0

\Rightarrow

x = - \frac{p}{2} \pm \sqrt{\frac{p^{3} + 9 p^{2} - 508}{12 (3 - p)}}

\Rightarrow \frac{p^{3} + 9 p^{2} - 508}{12 (3 - p)} ⩾ 0 \Rightarrow 3 < p ⩽ 5.8490 \dots

p \in Z \Rightarrow p = 4 \lor p = 5

p = 5 \Rightarrow x \notin Z

p = 4 \Rightarrow x = - 7 \lor x = 3

\Rightarrow (x, y) = (- 7, - 3) \lor (3, 7)

Commented by Lekhraj last updated on 14/Sep/20

T hanks

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