How many real roots has the equation? How many real roots (depends on the parametr $a$)  has the equation $x^{13}=a(x^{14}+1)?$
I ploted left and right sides and observed that for all $a \neq 0$ there is only one root. Is it true? 
Edit.  Correct typos.
 A: The equation is equivalent to
$P_a(x)=ax^{14}-x^{13} +a=0$
Check out Descartes' Rule of Signs:
https://en.wikipedia.org/wiki/Descartes'_rule_of_signs
It is very useful for this kind of problems. In this specific case, it says that $P_a(x)$ has exactly one zero, because 


*

*$P_a(x)$ has 1 change of signs for $a>0$ and 0 change of signs for $a<0$.

*$P_a(-x)$ has 0 change of signs for $a<0$ and 1 change of signs for $a>0$.

A: This is an answer to the original question: roots of $x^{13}=a(x^{13}+1)$.
Let us define $$f(x)=x^{13}-a^{13}-a=(1-a)x^{13}-a$$
Your function is strictly increasing if $1-a>0$, that is, $a>1$, and strictly increasing if $a<1$. In the first case, for $a>1$ we also have $$\lim_{x\to +\infty} f(x)=+\infty \qquad \lim_{x\to -\infty} f(x)=-\infty$$
so, applying Bolzano's theorem, there's a single point $c$ which satisfies $f(c)=0$. For $a<1$ is other way around, we have $\lim_{x\to -\infty} f(x)=+\infty$ and $\lim_{x\to -\infty} f(x)=+\infty$, but we arrive to the same conclusion.
For $a=1$, we get the function $f(x)=-1$, which obviously has no roots.
A: Note: This is in answer to OP's original question: How many real solutions to $x^{13}=a\left(x^{13}+1\right)$.
The equation of $x^{13}=\dfrac{a}{1-a}$ will have one real solution (either positive or negative when $1\ne a\ne0$. All $13$ solutions are equally spaced around a circle in the complex plane, thus none of the remaining $12$ will be real.
This is true in general for odd exponents of $x$. If the exponent is even, then there will be exactly two real solutions of the form $\pm r$.
A: Rewrite equation in form: $$a=\frac{x^{13}}{x^{14}+1}.$$
Now we will graph $y=\dfrac{x^{13}}{x^{14}+1}$. The reason is to discover how many roots the equation will have depending on values of parameter $a$. Notice that $y=a$ is a constant function and depending on it's position on the coordinate plane the equation will have a certain number of solutions.
Plotting non-constant function allows us to draw a conclusion about the number of roots of the equation:

*

*for $a\in\left(-\dfrac{\sqrt[14]{13^{13}}}{14}; 0\right)\cup\left(0; \dfrac{\sqrt[14]{13^{13}}}{14}\right)$ — two solutions;

*for $a\in\left\{-\dfrac{\sqrt[14]{13^{13}}}{14}, 0, \dfrac{\sqrt[14]{13^{13}}}{14}\right\}$ — unique solution;

*for $a\in\left(-\infty;-\dfrac{\sqrt[14]{13^{13}}}{14}\right)\cup\left(\dfrac{\sqrt[14]{13^{13}}}{14}; \infty\right)$ — no real solutions.
I saw this task in the collection of control papers, which were at the 3rd stage of the defense of scientific papers of the Junior Academy of Sciences of Ukraine. Where did you get this question from?

A: This is an answer to the original question :(
It is not necessary for such an issue to use analysis tools.
Two partiuclar cases. If $a=0$ then $x=0$. If $a=1$ then there is no solution.
Now, let us assume that $a \neq 0$ and $a\neq 1$; the initial expression is equivalent to :
$$x^{13}=b \ \text{with} \ b:=\dfrac{a}{1-a}.$$
Two cases:


*

*either $0<a<1 \iff \ b>0$ then $x=\sqrt[13]{b},$

*or $a$ outside $[0,1] \ \iff \ b<0$ then $x=-\sqrt[13]{b}.$
Conclusion: In all cases, there is one solution, but for $a=1$ where there are no solutions.
