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

Answered by MJS_new last updated on 13/Sep/20

$$x^3-y^3+9xy+127=0$$$$x,y,p\in \mathbb{Z}$$$$\mathrm{let}y=x+p$$$$(9-3p)x^2+(9p-3p^2)x-p^3+127=0$$$$x^2+px+\frac{p^3-127}{3(p-3)}=0$$$$\Rightarrow$$$$x=-\frac{p}{2}\pm \sqrt{\frac{p^3+9p^2-508}{12(3-p)}}$$$$\Rightarrow \frac{p^3+9p^2-508}{12(3-p)}⩾0\Rightarrow 3<p⩽5.8490...$$$$p\in \mathbb{Z}\Rightarrow p=4\vee p=5$$$$p=5\Rightarrow x\notin \mathbb{Z}$$$$p=4\Rightarrow x=-7\vee x=3$$$$\Rightarrow (x,y)=(-7,-3)\vee (3,7)$$

Commented by Lekhraj last updated on 14/Sep/20

$$T\mathrm{hanks}$$

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