Find two couples positive and intiger m, n which solve equation 2m^3=n^4
- Thread starter Jimmy44
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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 \(\displaystyle m=n=2\).
\(\displaystyle 2^{1}*m^{3}=n^{4}\) 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 \(\displaystyle m=n=2\).
D
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Can m = n?
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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?
\(\displaystyle n^4-2(kn)^3=0\)
\(\displaystyle n^3(n-2k^3)=0\)
Since we have \(\displaystyle 0<n\), what are we left with, and how can we use this to generate an infinite number of pairs?
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
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D
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As posted by Subhotosh Khan, begin by choosing any \(\displaystyle k\in\mathbb{N}\), then \(\displaystyle (m,n)=\left(k\left(2k^3\right),2k^3\right)=\left(2k^4,2k^3\right)\).
