Could someone help me with this?
If $m$ and $n$ are positive integers, then show that $$\frac{m}{ \sqrt n}+ \frac{m}{\sqrt[4]{n}} \neq 1$$.
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3Is it $$\frac{m}{\sqrt{n}} + \frac{m}{\sqrt[4]{n}} \neq 1;?$$ – Daniel Fischer Sep 01 '13 at 20:19
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That second term is supposed to be a fraction, right? – G Tony Jacobs Sep 01 '13 at 20:23
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This is obviously true as written, since $m\sqrt[4]{n}\geq 1$ so the left hand side is $>1$. So agreed with @DanielFischer, did you transcribe the problem incorrectly? – Thomas Andrews Sep 01 '13 at 20:29
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@ThomasAndrews I think that may be a mis-edit – Mark Bennet Sep 01 '13 at 20:39
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No, it should be right – Sep 01 '13 at 22:50
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The second term was the square root of the square root of n, which should be equal to the fourth root of n – Sep 01 '13 at 22:51
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$$\frac{m}{\sqrt n}+ \frac{m}{\sqrt[4]{n}} = 1 \implies \frac{m}{\sqrt n}\left(1+\sqrt[4]{n}\right) = 1 $$
Or $m(1+\sqrt[4]{n}) = \sqrt{n}$. Squaring, $m^2(1+\sqrt[4]{n})^2 = n$.
Suppose $n$ is not a perfect fourth power, so $r = \sqrt[4] n$ is irrational. Then we have $(1+r)^2 = N$, for some factor of $n$. So $r = \sqrt N -1$ and $N$ is not a perfect square.
However $r^4 = N^2+6N+1-4(1+N)\sqrt N= n$ makes $\sqrt N$ rational. Hence this case leads to contradiction.
So we need $n$ to be a perfect fourth power, say $n = k^4$. So $m^2(1+k)^2 = k^4$.
So $m(1+k) = k^2 \implies$ for all primes $p|(1+k)$, we must have $p | k$. This is not possible, hence we cannot have a solution.
Macavity
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3Would one not also need to show that $1+\sqrt[4]{n}\neq \sqrt{N}$ for integers $n,N?$ – L. F. Sep 01 '13 at 21:00
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