Solving Integer Cubes & Squares: 7k & 3k?

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Discussion Overview

The discussion revolves around the properties of integers that are both square and cube numbers, specifically focusing on their forms in relation to modulo 7 and modulo 3. Participants explore various approaches to understand these relationships, including references to Fermat's Little Theorem and the division algorithm.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant claims that an integer that is both a square and a cube must be of the form 7k or 7k+1, but expresses difficulty when working with (7k+2)^2 and (7k+3)^2.
  • Another participant suggests that using Fermat's Little Theorem can help establish that a^6 is either of the form 7k or 7k + 1, depending on whether a is divisible by 7.
  • There is a question about whether there is a similar modular characterization for cubic numbers as there is for square numbers modulo 3.
  • Some participants express a lack of familiarity with Fermat's Little Theorem and inquire about using the division algorithm instead.
  • A participant requests that if the discussion is related to homework, it should be posted in the appropriate homework forum.

Areas of Agreement / Disagreement

There is no consensus on the best approach to solving the problem, with some participants advocating for Fermat's Little Theorem while others seek alternative methods. The discussion remains unresolved regarding the applicability of these methods to cubic numbers.

Contextual Notes

Participants express varying levels of familiarity with number theory concepts, which may affect their contributions and understanding of the problem. The discussion includes references to specific mathematical techniques that may not be universally understood.

MAGICMATHS
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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 I am failing when i work (7k+2)^2, even (7k+3)^2....Can i interpret 7k+1 as 7k+(-1), i think i can't 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?
 
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MAGICMATHS said:
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 I am failing when i work (7k+2)^2, even (7k+3)^2....Can i interpret 7k+1 as 7k+(-1), i think i can't 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?

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?
 
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I don't know Fermat`s little Theory, i just started number theory to be honest ..is there a possibility using div alg pls?
 
MAGICMATHS said:
I don't know Fermat`s little Theory, i just started number theory to be honest ..is there a possibility using div alg pls?

Well, you could use the longer method presented here: http://www.mathyards.com/vb/showthread.php?691-Division-Algorithm-help !
 
Last edited by a moderator:
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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