<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] less beautiful
$$ Q^T D Q = H  $$
$$\left( 
\begin{array}{rrrr} 
1 & 0 & 0 & 0 \\ 
 -  \frac{ 1 }{ 2 }  & 1 & 0 & 0 \\ 
 -  \frac{ 1 }{ 2 }  &  -  \frac{ 1 }{ 3 }  & 1 & 0 \\ 
 -  \frac{ 1 }{ 2 }  &  -  \frac{ 1 }{ 3 }  &  -  \frac{ 1 }{ 2 }  & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrrr} 
2 & 0 & 0 & 0 \\ 
0 &  \frac{ 3 }{ 2 }  & 0 & 0 \\ 
0 & 0 &  \frac{ 4 }{ 3 }  & 0 \\ 
0 & 0 & 0 & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrrr} 
1 &  -  \frac{ 1 }{ 2 }  &  -  \frac{ 1 }{ 2 }  &  -  \frac{ 1 }{ 2 }  \\ 
0 & 1 &  -  \frac{ 1 }{ 3 }  &  -  \frac{ 1 }{ 3 }  \\ 
0 & 0 & 1 &  -  \frac{ 1 }{ 2 }  \\ 
0 & 0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrrr} 
2 &  - 1 &  - 1 &  - 1 \\ 
 - 1 & 2 & 0 & 0 \\ 
 - 1 & 0 & 2 & 0 \\ 
 - 1 & 0 & 0 & 2 \\ 
\end{array}
\right) 
$$
Back substitution says $t=0$  then $z=0$ then $y=0$ then $x=0.$   All that is going on is that the matrix I am calling $H$ is positive definite. With all real variables, each row in my $Q$ must indicate a zero vector ( when multiplied by $(x,y,z,t)^T \; \; \; \; $ )
</s>