This might be a trivial question, so, if it is, I will delete it as soon as I get an answer. I try to come up with conjectures once in a while. After all, the sleep of reason produces monsters (Francisco Goya, 1799). Recently, I came up with one (perhaps someone already did), and it is this:
Will the family of equations
$$a^n=k_1^{n-1}+k_2^{n-2}+\ldots+k_{n-2}^2,$$
always have at least one solution for each $n > 0$ and $k_i>0$?
Just so I'm even more clear, when $n = 8$, we have the following:
$$a^8=k_1^7+k_2^6+k_3^5+k_4^4+k_5^3+k_6^2.$$
I wrote a computer program and the cases $n = 3,4,5,6,7,8,9,10,11,12,13,14,15$ have at least one solution. I left out the cases $n = 1,2$ because one can obtain solutions by hand. So, has this ever been studied? I don't even know what to call this family of equations, so I can't just perform a Google search.
For $n=3$ we have the solution $1^2=1^3$, and for $n=4$ and $n=5$ we have the solutions \begin{array}{rrrrr} 28^2+&8^3\hphantom{+}&=&6^4\\ 2^2+&3^3+&1^4=&2^5 \end{array} Now suppose $n\geq5$ and that $k_2,k_3,\ldots,k_{n-1}$ are positive integers such that $$k_2^2+k_3^3+\ldots+k_{n-1}^{n-1}=2^n.$$ Then setting $k_n=2^n$ we have $$k_2^2+k_3^3+\ldots+k_{n-1}^{n-1}+k_n^n=2^n+2^n=2^{n+1},$$ which shows that we have a solution for $n+1$ as well. Hence by induction there exist solutions for all $n\geq3$.