Solve in non-negative integers the equation $$6^n+7=m^{k+2}$$
I found $n=0, m=2, k=1$. I don't know if there are any other solutions
Solve in non-negative integers the equation $6^n+7=m^{k+2}$
2
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number-theory
integers
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0For $1\le n\le 10^4$, $6^n+7$ is not a perfect power, so probably there are no more solutions. – 2017-01-25
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0$k$ has to be odd since $3$ is not a quadratic residue $\pmod{5}$. Apart from that, we have something similar to Catalan's conjecture (https://en.wikipedia.org/wiki/Catalan's_conjecture), and cyclotomic polynomials should be the key (by writing $6^n+6 = (m-1)(m^{k+1}+\ldots+1)$) – 2017-01-25
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0One can solve this using linear forms in logarithms to bound $k$ and then applying local arguments to rule out small values of the exponents. There might be an easy way to tackle this equation, but I don't see it off hand. – 2017-01-28