<s>[INST] Find $x$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$ Find all $x \in (-1, +\infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.
What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation: 
$(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} = (y^2 + 1)^{y - 2}$ and $y > 1$, which can then be rewritten as: 
$$ \left( \frac{y^2 + 1}{2y} \right)^{y - 2} - \left( \frac{y^2 - 1}{2y} \right)^{y - 2} = 1 $$
Also, my intuition is that $y = 4$ is the unique solution and initially I wanted to prove that this function of $y$ is injective, but it didn't work either.
Do you have any suggestion on how I can continue?
 [/INST] Hint:
$$(x^2+4x+5)^2-(x^2+4x+3)^2=2(2x^2+8x+8)=(2x+4)^2$$
$$\iff1=\left(\dfrac{2x+4}{x^2+4x+5}\right)^2+\left(\dfrac{x^2+4x+3}{x^2+4x+5}\right)^2$$
We have
$$\left(\dfrac{2x+4}{x^2+4x+5}\right)^x+\left(\dfrac{x^2+4x+3}{x^2+4x+5}\right)^x=1$$ 
WLOG $\dfrac{2x+4}{x^2+4x+5}=\cos t,\dfrac{x^2+4x+3}{x^2+4x+5}=\sin t$ with $0<t<\dfrac\pi2$ for $-1<x<\infty$
Clearly, $$(\cos t)^x+(\sin t)^x$$ is a decreasing function to forbid multiple solutions 
Generalization : $$\left(\dfrac{2x+4}{x^2+4x+5}\right)^y+\left(\dfrac{x^2+4x+3}{x^2+4x+5}\right)^y=1$$ 
$\implies y=2$
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