Thread: Another easy one View Single Post
2006-10-29, 15:29   #2
victor

Oct 2005
Fribourg, Switzerlan

22·32·7 Posts

Quote:
 Originally Posted by fetofs Could anyone enlighten me with the solution: Find all integer solutions of the equation $x^3-y^3=3(x^2-y^2)$ and explain why your answer is correct.
$\large{x^3-y^3=3(x^2-y^2)}$
$\large{(x-y)(x^2+x\cdot y+y^2)=3(x+y)(x-y)}$
$\large{x^2+xy+y^2=3(x+y)}$
therefore
$\large{x=\frac{sqrt{3-y}sqrt{3(y+1)}-y+3}{2}}$
or
$\large{x=\frac{-(sqrt{3-y}sqrt{3(y+1)}+y-3)}{2}}$

Last fiddled with by victor on 2006-10-29 at 15:29 Reason: thereforE<-