Problem 921

Let $d$ be the greatest common divisor of $m$ and $n$, i.e. $m = d m'$ and $n = d n'$, where $m'$ and $n'$ are relatively prime positive integers of different parity.
Now we need to prove that
$$\frac{3d^2 m'^2 + 5d^2 m' n'}{3d^2 n'^2 + d m' n'} = \frac{m'(3m' + 5n')}{n'(m' + 3n')}$$
is not a positive integer.
Note that $3m' + 5n'$ and $m' + 3n'$ are both odd.
If $m'$ is odd and $n'$ is even, the even $n'(m' + 3n')$ clearly cannot divide the odd $m'(3m' + 5n')$.
Otherwise, let the odd $k$ be the greatest common divisor of $3m' + 5n'$ and $m' + 3n'$. Then we have
$$\begin{aligned} k &\mid 3 \cdot (3m' + 5n') - 5 \cdot (m' + 3n'), & \text{ i.e. } k &\mid 4m', \\ k &\mid 1 \cdot (3m' + 5n') - 3 \cdot (m' + 3n'), & \text{ i.e. } k &\mid -4n', \end{aligned}$$
from which it follows that $k$ divides both $m'$ and $n'$, so $k = 1$ and both factors in
$$\frac{m'}{n'} \cdot \frac{3m' + 5n'}{m' + 3n'}$$
are irreducible fractions.
Since $m' + 3n' > m'$, the proof is finished.