Math contest proof equation problem Could someone help me with this?
If $m$ and $n$ are positive integers, then show that $$\frac{m}{
\sqrt n}+ \frac{m}{\sqrt[4]{n}} \neq 1$$.

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