# Math contest proof equation problem

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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Is it $$\frac{m}{\sqrt{n}} + \frac{m}{\sqrt[4]{n}} \neq 1\;?$$ – Daniel Fischer Sep 1 '13 at 20:19
That second term is supposed to be a fraction, right? – G Tony Jacobs Sep 1 '13 at 20:23
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 1 '13 at 20:29
@ThomasAndrews I think that may be a mis-edit – Mark Bennet Sep 1 '13 at 20:39
No, it should be right – Yadnarav3 Sep 1 '13 at 22:50

$$\frac{m}{\sqrt n}+ \frac{m}{\sqrt[4]{n}} = 1 \implies \frac{m}{\sqrt n}\left(1+\sqrt[4]{n}\right) = 1$$

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.

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Would one not also need to show that $1+\sqrt[4]{n}\neq \sqrt{N}$ for integers $n,N?$ – L. F. Sep 1 '13 at 21:00
@L.F. Covered explaining the other case. – Macavity Sep 1 '13 at 21:37
Surprisingly simple – Yadnarav3 Sep 1 '13 at 23:00