Is there an integer $m\geq 1$ such that $2^m+3^m$ is a perfect power?
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3$\begingroup$ The question is unclear (what is quantified how) and (at least the way it is phrased) it is not clear to me that this is a research level question. Please reformulate to explain what exactly you are asking. $\endgroup$– Max HornJun 27, 2011 at 13:02
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4$\begingroup$ I found the question perfectly clear actually: is there an integer $m \ge 1$ such that $2^m + 3^m$ is a perfect power? A quick numerical check gives no examples for $1 \le m \le 50000$. $\endgroup$– David LoefflerJun 27, 2011 at 13:08
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2$\begingroup$ David, yes, that seems like a plausible guess! But I'd still love if the original poster could clarify the question / quantization $\endgroup$– Max HornJun 27, 2011 at 13:13
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1$\begingroup$ For example, does the use of letter $p$ mean a prime? $\endgroup$– Gerald EdgarJun 27, 2011 at 13:43
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2$\begingroup$ Hi i'm the poster. Sorry i stated the question in a lax way. The question is exactly as DL understand it. $\endgroup$– EricJun 27, 2011 at 13:45
3 Answers
If you really wanted to prove this (and I'm afraid that I'm not sure why you would), you could invoke a Theorem of Darmon and Merel for $n=2$ and $3$, check that there are no solutions with $p \leq 5$, say, and then write down the usual $(n,n,n)$ Frey curve, assuming $n \geq 5$ is prime (which leads to a weight $2$, level $6$ cuspidal newform and hence the desired contradiction).
Of course, this is an absurdly big hammer for such a problem and likely something {\it much} simpler works.
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$\begingroup$ Using my notations: in that paper they assumed $p\geq 7$ is prime, NOT n (which is the exponent of $a$!). Are you sure you didn't mix the notations up? $\endgroup$– EricJun 27, 2011 at 16:16
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$\begingroup$ I think Mike did not mix the notation up. What he says is that the paper by Darmon and Merel implies that there are no solutions to your equation with $n=2$ and $n=3$. He then goes on to suggest that it suffices to consider $n$ prime (this is obvious: if $n$ is not prime, say $n=q\cdot m$, then rewrite $a^n$ as $(a^m)^q$), and then proposes a solution involving a Frey curve (which I don't claim to understand, as I don't know anything about Frey curves ;). $\endgroup$– Max HornJun 27, 2011 at 17:14
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$\begingroup$ The Frey curve corresponds to $$ 2^p \cdot 1^n + 3^p \cdot 1^n = a^n. $$ The ``machinery'' works for prime $n \geq 5$ in this case. Of course, it would work equally well for $$ 2^p+3^q=a^n $$ with $p \geq 4$. $\endgroup$ Jun 27, 2011 at 20:04
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$\begingroup$ Thanks! And to quote Gjergji Zaimi:"...(this problem) would amount to proving an effective version of Zsigmondy's theorem (where there is a Zsigmondy prime of exponent 1), which is at least as hard as FLT... I believe there are no simpler approaches" $\endgroup$– EricJun 28, 2011 at 6:29
By the Fermat theorem: n is not divisible by p.
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2$\begingroup$ Probably the case (2,3) is far easier than Wiles. $\endgroup$ Jun 27, 2011 at 13:42
I was hoping to see a post by Gjergji Zaimi on this question. My guess is he deleted whatever he might have had, from which I had hoped to learn something. So I will post a start of an elementary approach in the hopes that he or someone else will finish it.
Consider the case that p is an odd positive integer. Then a must be a multiple of 5, and either n is one, or else 2^p + 3^p is a multiple of 25, in which case p must be an odd multiple of 5 by considering the sum mod 25. So in this case a^n is a multiple of 3025 if n is not 1. This can probably be refined by looking at 2^p + 3^p mod 125, and hopefully considerations mod 11 may finish it off.
Now assume p=2q for some positive integer q. If n were even we could represent 3^p by (a - 2^q)(a+2^q), which would give 3^p = 2^(q+1) + 1, which is not solvable in integers p and q. So n must be odd, a^n must be 1 mod 8 for p sufficiently large, and so must a if n is not 1. Again, more work needs to be done here. It looks plausible to me that n must be 1 if a is an integer.
Gerhard "Email Me About System Design" Paseman, 2011.06.27
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2$\begingroup$ I deleted a post where I had written something trivial (case p prime, misread the question). I don't think this problem can be finished by elementary methods, though. It would amount to proving an effective version of Zsigmondy's theorem (where there is a Zsigmondy prime of exponent 1), which is at least as hard as FLT... $\endgroup$ Jun 28, 2011 at 3:26
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$\begingroup$ "It would amount to proving an effective version of Zsigmondy's theorem (where there is a Zsigmondy prime of exponent 1), which is at least as hard as FLT..." So has Mike got it right there? $\endgroup$– EricJun 28, 2011 at 4:09
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$\begingroup$ Yes, I believe there are no simpler approaches... $\endgroup$ Jun 28, 2011 at 4:48
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$\begingroup$ Thank you (as always) for your insights, Gjergji. Gerhard "Ask Me About System Design" Paseman, 2011.06.27 $\endgroup$ Jun 28, 2011 at 4:57