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

Jimmy44

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Feb 18, 2017
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1. Find two couples of positive and intiger m,n which solve equation
\(\displaystyle 2m^{3}=n^{4}\).
 
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 \(\displaystyle m=n=2\).
 
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. :D
 
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?
 
\(\displaystyle n^3(n-2k^3)=0\)
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
 
Last edited:
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
 
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 \(\displaystyle (m,n)=\left(k\left(2k^3\right),2k^3\right)=\left(2k^4,2k^3\right)\). :D
 
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