An algebra problem by Ojas Singh Malhi

$(x+y+z)^{2}$ =( $x^{2}+y^{2}+z^{2}$ )+2(xy+xz+xz)

9=3+2(xy+xz+xz)

Therefore,

xy+xz+xz=3

$x^{2}+y^{2}+z^{2}$ =3. (given)

Multiply by 2,

$2x^{2}+2y^{2}+2z^{2}$ =2xy+2yz+2zx

(As 3 = xy+xz+xz)

( $x^{2}$ -2xy+ $y^{2}$ )+( $y^{2}$ -2yz+ $z^{2}$ )+( $x^{2}$ -2zx+ $z^{2}$ )=0

$(x-y)^{2}$ + $(y-z)^{2}$ + $(z-x)^{2}$ =0

Therefore,

x=y=z

Therefore,

x=1 y=1 z=1 is the only solution.