Find all pairs $(x,y)$ of real numbers such that $ |x|+ |y|=1340$ and $x^{3}+y^{3}+2010xy= 670^{3}$ .
Problem
Source: JBMO 2010 Shortlist
Tags: algebra, equation, JBMO Shortlist
28.06.2017 12:42
$x^{3}+y^{3}+2010xy= 670^{3}\Leftrightarrow x^3+y^3+(-670)^3=3.(-670)xy\Leftrightarrow (x+y-670)(x^2+y^2+670^2-xy+670x+670y)=0$
03.01.2018 18:01
quangminh1173 wrote: $x^{3}+y^{3}+2010xy= 670^{3}\Leftrightarrow x^3+y^3+(-670)^3=3.(-670)xy\Leftrightarrow (x+y-670)(x^2+y^2+670^2-xy+670x+670y)=0$ If $x+y=670$ then.. $(x+y)^2-(|x|+|y|)^2 = (570)(2010)=2xy-2 \cdot |xy|$ and since $|xy|$ could not be defined at $xy$ (due to the contradiction) then $|xy|$ should be $-xy$ $\Rightarrow xy=1005 \cdot 285, x+y=670$ which has no real solutions. if $(x^2+y^2+670^2-xy+670x+670y)=0$ then $x=y=670$
27.05.2018 23:38
Above solution is wrong. From factorization we have 1/2×(x+y-670)×((x-y)^2+(x+670)^2+(y+670)^2)=0 If second bracket equals zero then solution is x=y=-670 If first bracket is equal to zero clearly one of numbers x and y is positive and other one is negative. WLOG x>0, y<0. x-y=1340 x+y=670 x=1005 y=-335 So solutions are (x,y)={(1005,-335),(-335,1005),(-670,-670)}