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. ?
How many positive integer solutions to $a^x+b^x+c^x=abc$?
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abstract-algebra
number-theory
diophantine-equations
exponentiation
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0Have you tried AM-GM to bound the solutions? – 2011-09-01
2 Answers
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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
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3This looks perfectly suited to be an answer rather than a comment. No "unfortunately" about that. (You will be able to write comments on arbitrary posts as soon as you reach 50 "reputation" points -- an "add comment" link will appear under each post then). – 2011-09-01
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0The $x=2$ 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. – 2011-09-02
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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.