solution set of linear equation Is there a system of linear equations with 3 variables and solution set $\{(a, b, c) | a^2 =  b\}$?
My answer is no but I'm not sure how to explain it.
Edit: can't I just say that if $ax+by+cz=ax+a^2y+cz=d$ is the equation then also $2ax+4a^2y+2cz=2ax+2a^2y+2cz+2a^2y=\color{red}d$. but if I multiply $ax+a^2y+cz=d$ by 2 then I get that $2ax+2a^2y+2cz=2d$. then if I substitute this into the previous equation I get that $2d+2a^2y=d\Rightarrow d=-2a^2y$ which also means that $(0, -2a^2, 0)$ is solution, but it's true only for $a=0$.
Is this correct way to prove it?  

There are some hidden assumptions here.  If your underlying field is $\mathbb{F}_2$, then the condition $a^2=b$ is equivalent to $a=b$, and in this case the system $\{a-b=0\}$ is a system that has the desired solution set.
If your underlying field is $\mathbb{R}$, then $(1,1,0)$ and $(3,9,0)$ are solutions, so by linearity $(2,5,0)$ is a solution, which is not in the desired solution set. [there is a fairly easy-to-prove theorem that states that the midpoint of two solutions is also a solution.]
