<s>[INST] If$x^2+y^2+z^2+t^2=x(y+z+t)$prove $x=y=z=t=0$ 
If
$$x^2+y^2+z^2+t^2=x(y+z+t)$$ Prove $x=y=z=t=0$

I added $x^2$ to both side of the equation:
$$x^2+x^2+y^2+z^2+t^2=x(x+y+z+t)$$
Then rewrite it as:
$$x^2+(x+y+z+t)^2-2(xy+xz+xt+yz+yt+zt)=x(x+y+z+t)$$
$$x^2+(x+y+z+t)(y+z+t)=2(xy+xz+xt+yz+yt+zt)$$
But it doesn't seem useful.
 [/INST] Note the beautiful identity $$y^2+z^2+t^2=\frac{1}{3}\left({(y+z+t)}^2+{(y-z)}^2+{(z-t)}^2+{(t-y)}^2\right)$$  thus we have $$x^2+\frac{1}{3}{(y+z+t)}^2+\frac{{(y-z)}^2+{(z-t)}^2+{(t-y)}^2}{3}=x(y+z+t)$$ $${\left (x-\frac{(y+z+t)}{2}\right)}^2+ \frac{{(y+z+t)}^2}{12}+\frac{{(y-z)}^2+{(z-t)}^2+{(t-y)}^2}{3}=0$$  which means $x=y=z=t=0$
(As its a sum of squares each term must be zero)
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