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This is a problem from the SUMS (Sydney University Mathematical Society) problems competition circa 1982 (open to undergraduates at all Australian universities and colleges). I couldn't prove it at the time, and having recently found it in a desk draw while having a clean up I still can't prove it.

I think it's relatively easy to show that if any such $m^n$ exists it must be of the form $5^n$. But it's not at all clear to me how to draw out a contradication to prove that no such $5^n$ exists for any $p$ which is prime. I suspect that Fermat's little theorem may play a role in there somewhere.

Many thanks for any guidance that you can offer.

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1 Answer 1

For $p=2,2^2+3^2=13$ which not of the form $n^m$ where $m,n\ge2$

So, we can start with $p\ge3.$

Now, $5\mid(2^p+3^p)$ for odd prime $p$

$2^p+3^p=2^p+(5-2)^p=2^p+\sum_{0\le r\le p}\binom p r5^{p-r}(-2)^r\equiv p\cdot5\cdot 2^{p-1}\not\equiv0 \pmod{5^2}$ if $p\ne 5$

If $p=5,2^5+3^5=32+243=275=5^2\cdot11$ so is not of the form $n^m$ where $m,n\ge2$

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Thank you for your elegant solution (as was no doubt intended by the original setter of the problem). –  CAM Nov 30 '12 at 12:25
    
For completeness should there be a term (-1)^r under the summation? Would the same argumnet work for any odd integer p >= 3 not divisible by 5, even those which are non prime? –  CAM Nov 30 '12 at 12:35
    
@CAM, sorry for the type error, rectified. –  lab bhattacharjee Nov 30 '12 at 12:36
    
@CAM, If $a+b=n, a^{2m+1}+(n-a)^{2m+1}\equiv (2m+1)na^{2m}\not\equiv0\pmod{n^2}$ if $(n,a(2m+1))=1$ or if $n\not\mid(2m+1)a^{2m}$ –  lab bhattacharjee Nov 30 '12 at 13:53

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