solve system equation: $ 2a^2 - 1 = b, 2b^2 - 1 = c, 2c^2 - 1 = a $ I have this system equation:

$$ 2a^2 - 1 = b $$ 
$$ 2b^2 - 1 = c $$
$$ 2c^2 - 1 = a $$

From system equation we see that $ a \neq 0 , b \neq 0, c \neq 0 $ , so :
$ 2a^2 - 1 \neq 0 =&gt; a \neq \sqrt{\frac 1 2} $ . 
Also we can write: $ b \neq \sqrt{\frac 1 2} $ And $ c \neq \sqrt{\frac 1 2} $ 

But I don't know if this is the answer or I need to find exact numbers that satisfy this system (btw, I found an answer and that's : $a = 1, b = 1, c = 1 $).
The given system of equations is symmetric, which means that by substituting the variables in the equations by any permutation of the set $\left (a,b,c \right )$ we get the same system all over again. Using that fact, equality $a=b=c$ must hold.  

Now simply by using $a=b$ from the first equation of the system we obtain:
$$2a^2-a-1=0$$ Using quadratic formula we get $a_{1}=1$ and $a_{2}=-\frac{1}{2}$.  

Therefore, the solution of the system is:

$$\left ( a,b,c \right )=\left \{ \left ( 1,1,1 \right ), \left ( -\frac{1}{2}, -\frac{1}{2},-\frac{1}{2} \right)  \right \}$$
The given system of equations is symmetric, which means that by substituting the variables in the equations by any permutation of the set $\left (a,b,c \right )$ we get the same system all over again. Using that fact, equality $a=b=c$ must hold.  

Now simply by using $a=b$ from the first equation of the system we obtain:
$$2a^2-a-1=0$$ Using quadratic formula we get $a_{1}=1$ and $a_{2}=-\frac{1}{2}$.  

Therefore, the solution of the system is:

$$\left ( a,b,c \right )=\left \{ \left ( 1,1,1 \right ), \left ( -\frac{1}{2}, -\frac{1}{2},-\frac{1}{2} \right)  \right \}$$