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More Div.Alg.

  1. Jul 22, 2012 #1
    I am trying to solve how an integer is simultaneously that is simultaneously a square and a cube number must be either of the form 7k or 7k+1 and im failing when i work (7k+2)^2, even (7k+3)^2.............Can i interpret 7k+1 as 7k+(-1), i think i cant but then i fail in many steps!! Also i know that any square number can be expressed as 3k or 3k+1...IS THERE SOMETHING SIMILAR FOR THE CUBIC PLEASE?
  2. jcsd
  3. Jul 22, 2012 #2


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    The easiest way is using Fermat's Little Theorem.

    Let a be any integer. If a is both a square and a cube, it is a sixth power.

    By Fermat's Little Theorem,

    [tex]a^7 \equiv a \pmod 7[/tex]

    Two cases:

    Case 1: [itex]7 \nmid a[/itex]

    Hence, [itex]a^6 \equiv 1 \pmod 7 \Rightarrow a^6 = 7k + 1, k \in \mathbb{Z}[/itex].

    Case 2: [itex]7 | a[/itex]

    Hence, [itex]a^6 \equiv 0 \pmod 7 \Rightarrow a^6 = 7k, k \in \mathbb{Z}[/itex].

    So [itex]a^6[/itex] is either of the form [itex]7k[/itex] or [itex]7k + 1[/itex] (QED).

    You can construct something exactly like this for the squares modulo 3, and indeed generalise it for any (p-1)th power modulo a prime p.

    Do you think the analogous statement for cubes would be true, given that 4 is composite?
    Last edited: Jul 22, 2012
  4. Jul 22, 2012 #3
    I dont know Fermat`s little Theory, i just started number theory to be honest ..is there a possibility using div alg pls?
  5. Jul 22, 2012 #4


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    Well, you could use the longer method presented here: http://www.mathyards.com/vb/showthread.php?691-Division-Algorithm-help [Broken]!
    Last edited by a moderator: May 6, 2017
  6. Jul 22, 2012 #5


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    MagicMaths, if this is homework or coursework, could you please post it in the appropriate homework forum (precalculus mathematics) in future, please?
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