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!
Exponential Diophantine Equations for Beginners
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number-theory
diophantine-equations
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0Do 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. – 2011-04-30
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0Yes I would like to know the simple obstructions to solutions and examples of it before studying very general and advanced theory – 2011-04-30
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51) Solve $a^b=b^{a^2}$. Hint: The equation is multiplicative, so use the prime factor decomposition. – 2011-04-30
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32) Solve $2^n=m^2+1$. Hint: What happens modulo 4? – 2011-04-30
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0@user9325, For the first if $d^i \in a$ and $d^j \in b$ then $ib = ja^2$. For $d=a$ this gives $a^2|b$ and for $d=b$ this gives $b|a^2$ therefore $a=b=1$. – 2011-04-30
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0@user9325, Reduction $\pmod 4$ forces $m = 4k + 1$ (unless $n=0$ which is one solution), multiplying that out and cancelling the two gives $2^{n-1} = 4m(2m+1) + 1$ which implies $n=1$. – 2011-04-30
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0@user9325, very nice problems! Especially the first. – 2011-04-30
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0(I used the notation $d^i \in a$ to mean that $i$ is the highest power of $d$ dividing $a$.. I don't know if there is a normal notation for it) – 2011-04-30
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1@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. – 2011-04-30
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0@user9325, Thanks for pointing out my mistake! I cannot find any way to solve this one – 2011-04-30
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2@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. – 2011-04-30
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0@quanta: The usual notation is: $p^a\parallel n$ iff $p^a\mid n$ and $p^{a+1}\nmid n.$ – 2012-04-04
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0There is a very interesting tool to solve this type of problem, the groups chosen. – 2017-07-23
1 Answers
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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: $S=[1^n+(n-1)^n]+[2^n+(n-2)^n]+...)$.