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 Post subject: perfect cubePosted: Sun, 18 Sep 2011 20:54:59 UTC
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Joined: Sun, 23 Mar 2008 10:55:47 UTC
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Find all for which is perfect cube of an integer.

edited

Last edited by mathemagics on Sun, 18 Sep 2011 21:03:45 UTC, edited 1 time in total.

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 Post subject: Re: perfect cubePosted: Sun, 18 Sep 2011 20:59:42 UTC
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mathemagics wrote:
Find for which is perfect cube of an integer.

This is easy just by inspection that x=4 works. Topic moved.

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 Post subject: Re: perfect cubePosted: Sun, 18 Sep 2011 21:03:11 UTC
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Joined: Sun, 23 Mar 2008 10:55:47 UTC
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Sorry! I mean find every for which is perfect cube

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 Post subject: Re: perfect cubePosted: Sun, 18 Sep 2011 21:08:20 UTC
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mathemagics wrote:
Sorry! I mean find every for which is perfect cube

OK, topic moved back.

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 Post subject: Re: perfect cubePosted: Sat, 29 Oct 2011 13:55:29 UTC
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x=2; x=4 ; x=6 are the only solutions

For x>6 x! + 3x - 9 == 5mod7

All cubes are 0mod7, 1mod7 or 6mod7

QED

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 Post subject: Re: perfect cubePosted: Sat, 29 Oct 2011 14:46:20 UTC
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NickMcG wrote:
x=2; x=4 ; x=6 are the only solutions

For x>6 x! + 3x - 9 == 5mod7

All cubes are 0mod7, 1mod7 or 6mod7

QED

Why does ? For example, .

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 Post subject: Re: perfect cubePosted: Sat, 29 Oct 2011 20:35:43 UTC
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Indeed, as outermeasure has pointed out, for any x>7, we are left with a linear congruence modulo 7, so we can make it be whatever we feel like by picking x properly.

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 Post subject: Re: perfect cubePosted: Sun, 30 Oct 2011 08:18:34 UTC
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Shadow wrote:
Indeed, as outermeasure has pointed out, for any x>7, we are left with a linear congruence modulo 7, so we can make it be whatever we feel like by picking x properly.

Indeed, if you want to use mod, you probably need some heavy analytic number theory machinery to help you pick a prime which is 1 (mod 3), for every sufficiently large , for which is not a cubic residue mod . Heuristically such prime should exist for all sufficiently large x
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If the cubic residues are uniformly distributed and independent, the probability of fixed x which doesn't have such prime p_x would be about , and so the expected number of such x is finite.

but you get into something very nasty very early on and AFAIK we don't know the behaviour of the cubic Gauss sum to say cubic residues are close enough to randomly distributed for the heuristic argument to work.

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