Find all real solutions to the equation$4^x+6^{x^2}=5^x+5^{x^2}$ Find all real solutions to the equation:$$4^x+6^{x^2}=5^x+5^{x^2}$$
Obviously $0$ and $1$ satisfy this equation, but here $x$ is a real number, and I do not know how to do the next step, in fact, I guess this equation is not in addition to $0$ and $1$ other than the root. But you know, I need a proof! Could anyone help me? Thanks a lot!

For any $x \in \mathbb{R}$, apply MVT to $t^{x^2}$ on $[5,6]$ and $t^x$ on $[4,5]$,
we find a pair of numbers $\xi \in (5,6)$ and $\eta \in (4,5)$ (both dependent on $x$) such that
$$6^{x^2} - 5^{x^2} = x^2 \xi^{x^2-1}\quad\text{ and }\quad
5^x - 4^x = x\eta^{x-1}$$
This implies
$$
{\tt LHS}-{\tt RHS} = (4^{x} + 6^{x^2}) - (5^x + 5^{x^2}) = x^2\xi^{x^2-1} - x\eta^{x-1}
$$
There are 3 cases we need to study:


*

*$x \in (1,\infty)$ - Both $x^2 - 1$ and $x - 1$ are positive, we have
$$\begin{align}
x^2\xi^{x^2-1} & \ge x^2\inf\{ t^{x^2-1} : t \in (5,6) \} = x^2 \left(\inf\{ t : t \in (5,6)\}\right)^{x^2-1} = x^2 5^{x^2-1}\\
x\eta^{x-1} & \le x\sup\{ t^{x-1} : t \in (4,5)\} = x\left(\sup\{ t : t \in (4,5)\}\right)^{x-1} = x 5^{x-1}
\end{align}\\
\implies
{\tt LHS} - {\tt RHS} \ge x^2 5^{x^2-1} - x5^{x-1} = (\underbrace{x}_{> 1}\underbrace{5^{x^2-x}}_{>1} - 1)\underbrace{x5^{x-1}}_{>0} > 0$$

*$x \in (0,1)$ - Both $x^2-1$ and $x-1$ are negative, we have
$$\begin{align}
x^2\xi^{x^2-1} & \le x^2\sup\{ t^{x^2-1} : t \in (5,6) \} = x^2 \left(\inf\{ t : t \in (5,6)\}\right)^{x^2-1} = x^2 5^{x^2-1}\\
x\eta^{x-1} & \ge x\inf\{ t^{x-1} : t \in (4,5)\} = x\left(\sup\{ t : t \in (4,5)\}\right)^{x-1} = x 5^{x-1}
\end{align}\\
\implies
{\tt LHS} - {\tt RHS} \le x^2 5^{x^2-1} - x5^{x-1} = (\underbrace{x}_{< 1}\underbrace{5^{x^2-x}}_{<1} - 1)\underbrace{x5^{x-1}}_{>0} < 0$$


*

*$x \in (-\infty,0)$ - We have 
$$x^2 \xi^{x^2-1} > 0 \land x \eta^{x-1} < 0
\quad\implies\quad
{\tt LHS} - {\tt RHS} = x^2\xi^{x^2-1} - x\eta^{x-1} > 0$$


Combine these 3 cases, we have ${\tt LHS} \ne {\tt RHS}$ for $x \in \mathbb{R} \setminus \{ 0, 1 \}$. 
Since we know $0, 1$ are roots of the equation at hand,
$0$ and $1$ are all the real roots of the equation.
