real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is 
Number of real solution of equation  $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is 

Plan
Put $x^2+6x+7=f(x)$. Then i have $f(f(x))=x$
For $f(x)=x$
$x^2+5x+7=0$ no real value of $x$
For $f(x)=-x$
$x^2+8x+7=0$ 
$x=-7,x=-1$
Solution given is all real solution
Help me please
 A: Let $x^2+6x+7=y$.
Thus, $y^2+6y+7=x$, which gives
$$x^2+6x+7-(y^2+6y+7)=y-x$$ or
$$(x-y)(x+y+7)=0.$$
Thus, $x=y,$ which does not give real solutions, or
$$y=-x-7,$$ which gives
$$x^2+6x+7=-x-7.$$
Can you end it now?
A: You are right that $f(f(x))=x$
This gives $f(x)=f^{-1}(x)$
Do you know anything about the relationship between a function and its inverse?
Try sketching both $y=f(x)$ and $y=f^{-1}(x)$ on the same axes...
A: Upon completing the square the equation $$(x^2+6x+7)^2+6(x^2+6x+7)+7=x$$
$$(x^2+6x+7)^2+6(x^2+6x+7)+9=x+2$$
$$[(x^2+6x+7)+3]^2=x+2$$
Which simplifies to $$[(x+3)^2+1]^2=x+2$$ which has no real solutions.
Upon comparison of derivatives for $x>-2$,  we see that  function on the LHS is always above the one on the RHS so they do not meet at any real values.
A: On completing the square, we have
\begin{align}
x= (x^2+6x+7)^2+6(x^2+6x+7)+7 &= ((x+3)^2-2)^2+6((x+3)^2-2)+7\\
&= (x+3)^4-4(x+3)^2+4+6(x+3)^2-5\\
&=  (x+3)^4+2(x+3)^2-1=  ((x+3)^2+1)^2-2 \geq 1-2=-1
\end{align}
Further, we alsoknow that $(x+3)^2$ is monotonically increasing for $x\geq -3$. Thus, we get that
\begin{align}
x= ((x+3)^2+1)^2-2\geq  ((-1+3)^2+1)^2-2 = 23.
\end{align}
Further, $((x+3)^2+1)^2-2\geq 3^34x+3^4-2\geq x, \forall x\geq 23$.
A: Set $y=x^2+6x+7,$ and note that the resulting system $$y^2+6y+7-x=0,\,\,\,x^2+6x+7-y=0$$ is symmetric. That is, performing the transformation $(x,y)\mapsto(y,x)$ leaves the system unchanged except for a permutation. Thus, for every solution $(x,y),$ it is always the case that $(y,x)$ is also a solution, where $x,y\in\mathrm R.$
It follows that this system can only have an even number of real solutions, which must be in pairs that are reflections of each other in the line $y=x,$ and they cannot both fall on this line because the given equation is a quartic with real coefficients, and as such cannot have an odd number of real roots.
But the parabolas defined by the system above are orthogonal, congruent parabolas. Thus, they can only intersect along the line $x=y.$ This contradicts the observation above that the points of intersection cannot fall on this line. Thus, there are no real solutions for the system, and therefore to the original quartic.
A: Rearrange:
$$(x^2+6x+7)^2+6(x^2+6x+7)+7=x \iff \\
((x+3)^2-2)^2+6((x+3)^2-2)+10=x+3 \iff \\
(x+3)^4+2(x+3)^2-(x+3)+2=0 \iff \\
(x+3)^4+(x+3-\frac12)^2+(x+3)^2+\frac74=0.$$
The LHS is positive, hence, there is no real solution.
