Find two couples positive and intiger m, n which solve equation 2m^3=n^4

[Jimmy44]

1. Find two couples of positive and intiger m,n which solve equation
$\displaystyle 2m^{3}=n^{4}$.

 

[Jimmy44]

I found only one couple. (m,n)=(2,2), because
$\displaystyle 2^{1}*m^{3}=n^{4}$ now we would have numbers in the 4th power on both sites. Then $\displaystyle 2^{1}=m^{1}$
so $\displaystyle 4=4$, so $\displaystyle m^{4}=n^{4}$ and $\displaystyle m=n=2$.

 

[Deleted member 4993]

1. Find two couples of positive and intiger m,n which solve equation
$\displaystyle 2m^{3}=n^{4}$.

Can m = n?

 

[Jimmy44]

I think yeah. In the exercise is written only that you have to find two couples of numbers. There is nothing about these numbers have to be different.

 

[MarkFL]

Suppose we let $\displaystyle m=kn$, then we may write:

$\displaystyle n^4-2(kn)^3=0$

$\displaystyle n^3(n-2k^3)=0$

Since we have $\displaystyle 0<n$, what are we left with, and how can we use this to generate an infinite number of pairs?

 

[Jimmy44]

$\displaystyle n^3(n-2k^3)=0$

Thanks, do you mean that
$\displaystyle n=2k^3$? Because $\displaystyle n^3$ can't equal 0 (because then $\displaystyle n=0$ and this is contradictory with text of exercise).
But how to find $\displaystyle n$,$\displaystyle m$ and $\displaystyle k$ which solve this equation?t

 

[Deleted member 4993]

Thanks, do you mean that
$\displaystyle n=2k^3$? Because $\displaystyle n^3$ can't equal 0 (because then $\displaystyle n=0$ and this is contradiction with text of exercise).
But how to find $\displaystyle n$,$\displaystyle m$ and $\displaystyle k$ which solve this equation?

I'll do one for you.

Suppose we let [FONT=MathJax_Math-italic]m[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math-italic]k[/FONT][FONT=MathJax_Math-italic]n[/FONT]

Then for
k=1→

n=2k³→

n=2→

m=2

 

[MarkFL]

Thanks, do you mean that
$\displaystyle n=2k^3$? Because $\displaystyle n^3$ can't equal 0 (because then $\displaystyle n=0$ and this is contradictory with text of exercise).
But how to find $\displaystyle n$,$\displaystyle m$ and $\displaystyle k$ which solve this equation?t

As posted by Subhotosh Khan, begin by choosing any $\displaystyle k\in\mathbb{N}$, then $\displaystyle (m,n)=\left(k\left(2k^3\right),2k^3\right)=\left(2k^4,2k^3\right)$.