Solutions of $ 5^x+5^{x^2}=4^x+6^{x^2} \quad \left(x\in \mathbb{R}\right)$ For equation $$ 5^x+5^{x^2}=4^x+6^{x^2} \quad \left(x\in \mathbb{R}\right)$$
is there any nontrivial solution?
Easy to find $x=0,1$ are the trivial solutions, and also, easy to figure out that $ x>1 $ or $ x<0 $ would not satisfy the equation.
But I found it hard to deal with the $ \left(0,1\right) $ case, and I haven't got any idea yet.
Any Advice Would Be Greatly Appreciated.
 A: We prove there is no solution in $(0,1)$. The equation is equivalent to
$$\iff 5^x\left(1-\left( \frac{4}{5} \right)^x \right) = 5^{x^2}\left(\left( \frac{6}{5} \right)^{x^2} -1\right)$$
$$\iff 5^{x-x^2}\left(1-\left( \frac{4}{5} \right)^x \right) = \left( \frac{6}{5} \right)^{x^2} -1 \tag{1}$$
We observe that, for $x \in (0,1)$

*

*The LHS of $(1)$ is greater than $1-\left( \frac{4}{5} \right)^x$

*The RHS of $(1)$ is smaller than $\left( \frac{6}{5} \right)^{x} -1$
Then from $(1)$, we deduce that
$$(1) \implies 1-\left( \frac{4}{5} \right)^x < \left( \frac{6}{5} \right)^{x} -1 \implies \left( \frac{6}{5} \right)^{x}+\left( \frac{4}{5} \right)^x > 2  \tag{2} $$
(The equality doesn't occur for $x \in (0,1)$)
Study the function $f(t) = t^x$ for $t\in \Bbb R^+$. We have $f''(t) = x(x-1)t^{x-2}<0$ for $x\in (0,1)$. Applying the Jensen's inequality, we obtain that
$$\frac{1}{2}  \left(  f\left(\frac{4}{5}  \right)+f\left(\frac{6}{5}  \right)  \right)  \le f\left(\frac{1}{2}  \left(\frac{4}{5} +\frac{6}{5}  \right)  \right) =1$$
or
$$\left( \frac{6}{5} \right)^{x}+\left( \frac{4}{5} \right)^x \le 2  \tag{3}$$
From $(2)$ and $(3)$, we deduce that there is no solution for $x \in (0,1)$.
Q.E.D
