Is there a way to see that every solution of $ y^{r_1}-1=cy^{r_2}$ is constant without differentiating? Suppose $y:\mathbb{R} \to \mathbb{R}$ is a positive smooth function, satisfying the equation $$ y^{r_1}-1=cy^{r_2},$$
where $c,r_1,r_2 \in \mathbb{R}$ are constants, $r_1,c \neq 0$.
Is there a way to prove that $y$ must be constant without using derivatives?
(Here is an approach which does use derivatives:)
By differentiating, we get
$$ r_1y^{r_1-1}y'=cr_2y^{r_2-1}y'.$$
Suppose by contradiction there exist a point $x$, where $y'(x) \neq 0$. Then $y' \neq 0$ in a neighbourhood of $x$. Now, at every point where $y' \neq 0$, we get  $ r_1y^{r_1-1}=cr_2y^{r_2-1} \Rightarrow y^{r_1-r_2}=c\frac{r_2}{r_1}$. So, if $r_1 \neq r_2$ $y$ is constant. Otherwise, $c=1$, and the original equation becomes $y^{r_1}-1=y^{r_1}$ which is a contradiction.
Hence, $y$ is constant in a neighbourhood of $x$, so $y'(x)=0$.
 A: Let 
$$f(x)=-0.25x^5+x^4-1$$
This fnction has two positive roots, as can bee seen, since:


*

*$f(0)=-1<0$,

*$f(2)=7>0$,

*$f(4)=-1<0$ and

*$f$ is continuous


Let $x_1,x_2$ be these roots and let:
$$y(x)=\left\{\begin{array}{ll}x_1 & x\in\mathbb{R}\setminus\mathbb{Q}\\x_2 &  x\in\mathbb{Q}&\end{array}\right.$$
Now, it is clear that, for $r_1=4$, $r_2=5$ and $c=\frac{1}{4}$ we have:
$$y^{r_1}-1=cy^{r_2}$$
Also $y>0$.
So, smoothness is needed as a hypothesis, as we found a non-smooth $y$ that is not constant and, yet, satisfies the given equality.
Hope this helped! :)
A: A try for a (not simple...) solution. First, I suppose that $r_1r_2c\not =0$, and in addition that $r_1\not=r_2$. Let $F(t)=1+ct^{r_1}-t^{r_2}$. Suppose that there exist an interval $I=[a,b]$, with $0<a<b$ such that $F(t)=0$ for all $t$ in $I$. Choose  $\alpha$ and a $\beta>0$ such that $\log a<\alpha<\alpha+\beta<\alpha+2\beta<\log b$. Put $t_k=\exp(\alpha)\exp(k\beta)$ for $k=0,1,2$. Then $F(t_k)=0$ and the linear system $x_1+x_2 \exp(r_1\beta)^k+x_3\exp(r_2\beta)^k=0$ $k=0,1,2$ has a non trivial solution $(1, c\exp(r_1\alpha), -\exp(r_2\alpha))$. Hence the determinant of the system must be $0$ (it is a Vandermonde  determinant). We get that $(\exp(r_1\beta)-1)(\exp(r_2\beta)-1)(\exp(r_1\beta)-\exp(r_2\beta))=0$, a contradiction.
Now your function $y(x)$ is such that $F(y(x))=0$. If your continuous $y$ is not constant, then its range is a non trivial  interval, contradiction by the above. It remains to look at the excluded cases. 
