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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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Do you have anything particular in mind that you want to learn about? There are several kinds of simple obstructions to solutions, but this kind of result is not very general at all. And somewhat more general results, say Catalan's conjecture, are still special and difficult. –  Phira Apr 30 '11 at 9:01
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1) Solve $a^b=b^{a^2}$. Hint: The equation is multiplicative, so use the prime factor decomposition. –  Phira Apr 30 '11 at 13:58
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2) Solve $2^n=m^2+1$. Hint: What happens modulo 4? –  Phira Apr 30 '11 at 13:58
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@quanta Your argument for the first problem is incorrect. It works only for $d$ prime, so if you apply it to $a$, you have only proved that $a$ is 1 or not prime. –  Phira Apr 30 '11 at 14:25
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@quanta Note that $i/j$ does not depend on $d$, that is, either all exponents of primes in $a$ are smaller than the exponents in $b$ or the reverse, so you either have $a=bk$ or $b=ak$. You can substitute the two cases, simplify and iterate the argument if necessary. –  Phira Apr 30 '11 at 20:44
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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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