I need a direction, I was given a hint: the claim is somehow how related to another claim, for 2 consecutive even numbers one is divisible by 4 the other isn't.

  • $\begingroup$ Can three?????? $\endgroup$ – barak manos Oct 29 '15 at 7:36
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    $\begingroup$ @barakmanos: the hint cannot be used for three numbers. $\endgroup$ – Yves Daoust Oct 29 '15 at 7:37
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    $\begingroup$ If they are first powers you can have as many as you like. $\endgroup$ – Mark Bennet Oct 29 '15 at 7:49
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    $\begingroup$ Even two consecutive numbers can't be powers, except for $8,9$. This is because of Mihăilescu's theorem. $\endgroup$ – user236182 Oct 29 '15 at 9:41
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    $\begingroup$ $2 = 2^1 , 3=3^1, 4=2^2, 5=5^1$ $\endgroup$ – Michael Stocker Oct 29 '15 at 10:55

Among the four consecutive numbers, one must be of the form $4n+2=2(2n+1)$. As the multiplicity of $2$ in its prime decomposition is $1$, this number cannot be a power.

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    $\begingroup$ However, $2,3,4,5$ works if you count primes as powers of primes. $\endgroup$ – John Dvorak Oct 29 '15 at 13:17
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    $\begingroup$ @JanDvorak: allowing first powers is of little interest. $\endgroup$ – Yves Daoust Oct 29 '15 at 13:45
  • $\begingroup$ I know Mihăilescu's theorem, but this thread makes me wonder if there is an elementary way to see that the three consecutive integers $4n-1$, $4n$ and $4n+1$ can never be simultaneously perfect powers. Becuase four consecutive are very easy to rule out, and two consecutive were a tough problem for a long time. $\endgroup$ – Jeppe Stig Nielsen Oct 29 '15 at 14:33

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