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What would be some exponential Diophantine equations for the beginner to solve (which can demonstrate the techniques?) especially good if there are hints! Thank you very much!

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    There is$a$very interesting tool to solve this type of problem, the groups chosen.2017-07-23

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The posed problem is tightly connected with FLT, which here is not examined. But it is it's a pity! However,… If Fermat’s equality exists, then in the numeration system with the prime base n>2 next-to-last digits in numbers $1^n$, $2^n$,...$(n-1)^n$ are equal to 0 and, therefore, the two-digit end of the number $S=1^n+2^n+...+(n-1)^n$ is equal to the sum of the arithmetical progression S'=1+2+...+(n-1), i.e. is equal to the number $d0$, where the digit $d$ is not zero. That contradicts the direct calculation of the end of the number S (it is equal to 00, which is evident when grouping the terms of the sum $S$ into the pairs: $S=[1^n+(n-1)^n]+[2^n+(n-2)^n]+...)$.

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