# Proof that equation does not have real roots

For polynom $$f(x)=ax^2+bx+c$$ equation $$f(x)=x$$ has no real solutions. Prove that equation $$f(f(x))=x$$ also does not have does not have real solutions

Can someone explain solution to me? Why?

If equation $$f(x)=x$$ has no real solutions than it is either $$f(x)>0$$ and $$a>0$$ or it is $$f(x)<0$$ and $$a<0$$. In first case $$f(f(x))>f(x)>x$$ or in second case $$f(f(x)) for no real number $$x$$ it can not be $$f(f(x))=0$$

• Your assumption about $f$ is wrong: consider $f(x)=(x+2)^2-1$. Dec 23, 2018 at 19:59
• The quoted solution is incorrect. Dec 23, 2018 at 20:00
• @John_Wick Your conclusion is wrong. Consider $f(x)=(x+2)^2-1$, then $f(f(-0.9)),f(-0.9)$. Dec 23, 2018 at 20:05

The quoted solution should read

If equation $$f(x)=x$$ has no real solutions than it is either $$f(x)>\color{red}x$$ and $$a>0$$ or it is $$f(x)<\color{red}x$$ and $$a<0$$. In first case $$f(f(x))>f(x)>x$$ or in second case $$f(f(x)) for no real number $$x$$ it can not be $$f(f(x))=\color{red}x$$.

If $$a>0$$ then $$f(x)=ax^2+bx+c>x$$ for sufficiently large $$x$$. And as $$f(x)=x$$ has no real root then $$f(x)>x$$ for all $$x$$ (otherwise there will be a real root). Hence $$f(f(x))>f(x)>x$$ for all $$x$$. So, $$f(f(x))=x$$ has no real root. Similarly for $$a<0.$$

Let

$$g(x)=f(x)-x=ax^2+(b-1)x+c$$

no real roots implies that

$$(b-1)^2-4ac<0$$ and $$(\forall x\in \Bbb R)\;\; g(x) \text{ has a constant sign} :$$ $$(+ \text{ if } a>0 \text{ and } - \text{ if } a<0)$$.

thus $$f(f(x))-x=$$ $$f(f(x))-f(x)+f(x)-x=$$ $$g(f(x))+g(x)$$ will have a constant sign and no root.