Consider the diophantine equation
For distinct strict positive integers $A,B,C,D,E$ :
$$A^{15} + B^5 + C^5 + D^5 = E^{10}$$
One solution is
$$3^{15} + 84^5 + 110^5 + 133^5 = 12^{10}$$
Are there any others ?
Is it true that $A,B,C,D$ can not all be primes ?
I tried modular arithmetic but with no results.
I got fascinated by this because the equation has many high powers but only $5$ terms.
Best Answer
Let $a^5+b^5+c^5+d^5=e^5$. Then $(ae)^5+(be)^5+(ce)^5+(de)^5=e^{10}$, and thus $(ae)^{15}+(a^2be^3)^5+(a^2ce^3)^5+(a^2de^3)^5=(ae^2)^{10}$.
Also, note that if $(A,B,C,D,E)$ is a solution of given equation, then $(k^2A, k^6B, k^6C, k^6D, k^3E)$ is also the solution.
Currently we we only know two positive solutions of $a^5+b^5+c^5+d^5=e^5$, which are \begin{align*} &27^5+84^5+110^5+133^5=144^5\quad \text{(L. J. Lander, T. R. Parkin, 1966)}\\ &55^5+3183^5+28969^5+85282^5=85359^5\quad \text{(J. Frye, 2004)} \end{align*}
For the first solution, we have $A=2^4\times 3^5$, $B=2^{14}\times 3^{13}\times 7$, $C=2^{13}\times 3^{12}\times 5\times 11$, $D=2^{12}\times 3^{12}\times 7\times 19$, $E=2^8\times 3^7$. Now we may take $k=2^2\times 3^2$, and this gives your result.
For the second solution, we have $A=3\times 5^2\times 11\times 5693$ (at this point $k=1$ or $5$), $B=3^4\times 5^5\times 11^2\times 1061\times 5693^3$ (at this point $k=1$), and so on, which gives another "minimal" counterexample which is not related to your solution (but which have a nontrivial common factor). This gives \begin{equation*} 4694745^{15}+5988388578529986097425^5+54501297119520944786775^5+160446671301977466025950^5=400738738455^{10}. \end{equation*}
See https://www.wolframalpha.com/input?i=4694745%5E15%2B5988388578529986097425%5E5%2B54501297119520944786775%5E5%2B160446671301977466025950%5E5-400738738455%5E10.