<s>[INST] 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
 [/INST] 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, $((x+3)^2+1)^2-2\geq 3^34x+3^4-2\geq x, \forall x\geq 23$.
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