Find a positive integer 'm' such that m/2 is a perfect square and m/3 is a perfect cube

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Discussion Overview

The discussion revolves around finding a positive integer 'm' such that m/2 is a perfect square and m/3 is a perfect cube. Participants explore the mathematical properties and implications of the problem, including prime factorization and potential solutions.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant suggests considering the prime decomposition of m to approach the problem.
  • Another participant proposes a specific solution, stating that M=34·23=648 satisfies the conditions.
  • A later post shifts the topic to a different question regarding the construction of an equilateral triangle using lattice points, indicating a potential misunderstanding of the original problem.
  • Further, a participant argues that constructing such a triangle is impossible if one side is parallel to one of the coordinate axes, providing a height formula to support their reasoning.

Areas of Agreement / Disagreement

There is no consensus on the original problem regarding the integer 'm', as the discussion includes a proposed solution but does not explore further validation or alternative solutions. The later question about the equilateral triangle introduces a separate debate without resolution.

Contextual Notes

The discussion contains assumptions about the properties of integers and geometric constructions, but these are not fully explored or resolved. The transition to the triangle question introduces additional complexity without a clear connection to the initial problem.

bomba923
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"Find a positive integer 'm' such that m/2 is a perfect square and m/3 is a perfect cube."

How to solve it?
 
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consider m's prime decomposition
 
The smallest one can find is

[tex]M=3^{4}\cdot 2^{3}=648[/tex]

Daniel.
 
Prime factorization::thanks

Now just one more question: Is it impossible to construct an equilateral triangle using only lattice points?---(i.e., points with integer coordinates)
My work says no---but can it be done?
 
Nope,if one of the sides is || with one of the axis of coordinates...The height of the triangle is

[tex]h=\frac{l\sqrt{3}}{2}[/tex]...(that justifies it)...

As for the general case,it is a system of 3 equations with 6 unknowns...

Daniel.
 

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