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

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To find a positive integer 'm' such that m/2 is a perfect square and m/3 is a perfect cube, the prime factorization approach leads to the solution m = 648, which is derived from the factors 3^4 and 2^3. The discussion also touches on the impossibility of constructing an equilateral triangle using only lattice points, confirming that it cannot be done if one side is parallel to an axis. The height of such a triangle is given by the formula h = (l√3)/2, which supports this conclusion. The problem of lattice points involves a complex system of equations, indicating a deeper mathematical challenge. Overall, the thread explores both number theory and geometric constraints.
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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

M=3^{4}\cdot 2^{3}=648

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

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

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

Daniel.
 

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