# How many positive integer solutions to $a^x+b^x+c^x=abc$?

How many positive integer solutions are there to $a^{x}+b^{x}+c^{x}=abc$? (e.g the solution $x=1$, $a=1$, $b=2$, $c=3$). Are there any solutions with $\gcd(a,b,c)=1$? Any solutions to $a^{x}+b^{x}+c^{x}+d^{x}=abcd$ etc. ?

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Have you tried AM-GM to bound the solutions? – Bonanza Sep 1 '11 at 17:12

Unfortunately, I don't know how to make comments here, would be better to move it there. Anyways, consider for example, $a^x+b^x+c^x=abc$ with $a\le b\le c.$ Clearly, $c^x<LHS\le c^3,$ therefor $x<3.$ Case $x=1$ is fairly simple as you can bound $ab\le 3$ (following the same lines). So let's consider the case $x=2.$ Our equation is quadratic with respect to $a,$ so it has an integer solution iff $D=(c^2-4)b^2-4c^2=x^2.$ This equation has infinitely many solutions even for some fixed values of $c.$ Take. for example, $c=3.$ Then, $x^2-5b^2=-36,$ has infinitely many solutions, since it is Pell type equation with one solution being $x=12,$ $b=6.$ By considering equation modulo $3,$ we get $(a,b,c)\ne 1.$
The $x=2$ case is reminiscent of the Markov equation (q.v.), $a^2+b^2+c^2=3abc$. Markov is a special case of the Hurwitz equation, $x_1^2+\cdots+x_n^2=ax_1\cdots x_n$, for which see D12 in Guy, Unsolved Problems In Number Theory. – Gerry Myerson Sep 2 '11 at 6:25
The $x=1$ problem, $$a_1+a_2+\cdots+a_k=a_1a_2\cdots a_k$$ is discussed at some length at D24 in Guy's Unsolved Problems In Number Theory. There is always the solution $a_1=2$, $a_2=k$, $a_3=a_4=\cdots=a_k=1$. Various people have shown that $k=2,3,4,6,24,114,174$, and $444$ are the only $k\le1440000$ for which there is exactly one solution. See the book for more questions, references, etc.