Consider the following polynomials \begin{equation*} \begin{cases} f_1(x) = x^{m_1} - x^{n_1} - 1, \ &m_1>n_1>0 \\ f_2(x) = x^{m_2} - x^{n_2} - 1, \ &m_2>n_2>0. \end{cases} \end{equation*} Problem: For which values of $m_1,n_1,m_2,n_2$ the system of polynomial equations $f_1 = f_2 =0$ has a real solution $x_0>1$?
Some considerations:
Observe that $f_i(1) = -1$ and $\lim_{x \to +\infty} f_i(x) = +\infty$; hence, by the intermediate value theorem, $f_i$ has a real solution greater than unity. Additionally $f_i'(x) = m_ix^{m_i-1} - n_ix^{n_i-1} >0, \ \forall x>1$, from which these polynomials are strictly increasing. Thus such a root is unique.
So the problem may be reformulated as whether the unique real roots greater than unity of $f_1$ and $f_2$ coincide.
An example of solution is given by $m_1 = 5$, $n_1 = 4$, $m_2 = 3$, $n_2 = 1$, for which the polynomials $f_1$ and $f_2$ share the root $\lambda = \frac{\sqrt[3]{9-\sqrt{69}}+\sqrt[3]{9+\sqrt{69}}}{\sqrt[3]{2}3^{2/3}} \approx 1.325$. This example can be extended to a family of solutions given by $m_1 = 5k$, $n_1 = 4k$, $m_2 = 3k$, $n_2 = k$, for any $k \in \mathbb{N}$, whose roots are $\lambda^{1/k}$, and I verified numerically that these are the only solutions for polynomials with degree up to 100.
I would like to know if the above family of solutions is actually the only solution for this problem.
Thanks in advance for any help or insight about this problem.