Finding nontrivial solutions to a system of equations Suppose I have two equations for real numbers
$$
a^2+b^2+c^2+d^2+e^2+f^2=1
$$
$$
af-be+cd=0
$$
I would like to find a solution $(a,b,c,d,e,f)$ such that none of its entries are zero and all entries are distinct, how do I go about doing that?
 A: By adding (subtracting) twice the second equation to (from) the first, your system is equivalent to \begin{eqnarray*}(a+f)^2+(b-e)^2+(c+d)^2&=&1,\\
(a-f)^2+(b+e)^2+(c-d)^2&=&1.
\end{eqnarray*}
Thus $(a\pm f,b\mp e,c\pm d)$ can be thought of as points on the unit 2-sphere, and so can be parametrized by 
\begin{eqnarray*}(a+f,b-e,c+d)&=&(\cos\theta,\sin\theta\sin\phi,\sin\theta\cos\phi),\\
(a-f,b+e,c-d)&=&(\cos\theta',\sin\theta'\sin\phi',\sin\theta'\cos\phi').
\end{eqnarray*}
This yields $a=(\cos\theta+\cos\theta')/2$ etc, giving a representation of the 4-parameter ($\theta,\theta',\phi,\phi')$ solution. Pick a value for each parameter to get a particular solution. 
A: From your equetions one has $(a+f)^2+(b-e)^2+(c+d)^2=1$.
If you want to get some solution, set, for example, $a+f=c+d=2(b-e)$, then one has $b-e=\pm 1/3$  etc.
A: With two equations in six unknowns, you should be able to pick four of them and solve for the other two.  Because of the sum of squares, the numbers need to be small, in particular less than 1 in absolute value.  It is easier if you are solving for two variables that aren't multiplied  together in the second equation, so let us imagine we will solve for $a$ and $b$.  Then let $c^2+d^2+e^2+f^2=g, cd=h$, both of which will be known when we pick the four variables.  Now the equations are $$a^2+b^2=1-g \\ af-be=-h$$ which we can solve by substitution.  $$a=\frac{be-h}f\\ \frac {(be-h)^2}{f^2}+b^2=1-g\\(e^2+f^2)b^2-2ehb+h^2+gf^2-f^2=0$$ which is a quadratic we can solve by the usual formula.  I just took $c=0.1,d=0.2,e=0.3,f=0.4$ and found $a \approx 0.469422, b\approx 0.692563$ and $a \approx -0.53342, b\approx -0.64456$ as solutions.
