Given $$ x-y=2 $$ , $$ x^2+y^2=4 $$ , find the value of $$ x^{2002}+y^{2002} $$

Question

Uniteasy Admin · Accepted Answer

$$\because x-y=2 $$n
Squaring both sides,
$$(x-y)^2 = 4$$
$$	herefore  x^2+y^2-2xy = 4$$
$$\because x^2+y^2=4 $$
$$xy=0$$
$$x=0 $$ or $$y=0$$
if $$x=0, y=-2$$
if $$y=0,x=2$$
Either cases yields the same result,
$$ x^{2002}+y^{2002} $$
$$ = 2^{2002} $$

Question

Answer