I need help about this:
I a,b,c are real numbers such that $\displaystyle (b-1)^2-4ac<0$ , proof that system
$\displaystyle ax^2+bx+c=y\\ ay^2+by+c=z \\ az^2+bz+c=x$
don't have any solution in set of real numbers.
I think that it is posible to resolve by prooving that x=y=z (I dont know how to ) , and than will be (b-1)^2-4ac<0 (it is supposed by problem). So, equation(s) dont have real solutions.
Sorry for my bad english. I wish you understood my idea.
after a full day,my idea:
your equations:
$\displaystyle ax^2+bx+c-y=0$
$\displaystyle ay^2+by+c-z=0$
$\displaystyle az^2+bz+c-x=0$
summing them up we get $\displaystyle ax^2+bx+c-y+ay^2+by+c-z+az^2+bz+c-x=0$
rearranging the terms, $\displaystyle ax^2+bx-x+c+ay^2+by-y+c+az^2+bz-z+c=0$
or, $\displaystyle \bigg(ax^2+(b-1)x+c \bigg)+\bigg(ay^2+(b-1)y+c \bigg)+\bigg(az^2+(b-1)z+c \bigg)=0$
now by completing the squares we try to make the expression in the brackets a square.In general case:
$\displaystyle am^2+bm+c=a(m-h)^{2}+k$, where $\displaystyle h=\frac{-b}{2a}$ and $\displaystyle k=c-\frac{b^2}{4a}$
using the same principles we get,
$\displaystyle \bigg(a(x+\frac{b-1}{2a})^2+c-\frac{(b-1)^2}{4a}\bigg)+\bigg(a(y+\frac{b-1}{2a})^2+c-\frac{(b-1)^2}{4a}\bigg)+\bigg(a(z+\frac{b-1}{2a})^2+c-\frac{(b-1)^2}{4a}\bigg)=0$
simplifying we get,
$\displaystyle a[(x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2]+3(\frac{4ac-(b-1)^2}{4a})=0$
or,$\displaystyle a[(x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2]=-3(\frac{4ac-(b-1)^2}{4a})$
dividing both sides by $\displaystyle a$,
$\displaystyle (x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2+(x+\frac{b-1}{2a})^2=-\frac{3}{4}.\frac{4ac-(b-1)^2}{a^2}$
now, $\displaystyle (b-1)^2-4ac<0$ (given) $\displaystyle \implies 4ac-(b-1)^2>0$ also $\displaystyle a^2>0$ therefore, $\displaystyle \frac{4ac-(b-1)^2}{a^2}>0$ and thus the right hand side of the equation is a negative number, while the left hand side is the sum of squares and should always be positive.
This is not possible for any real number solution, and hence we conclude no real solution exist.