Ping sent this solution from Thailand.

(1) If $5^{m} = 4^{m} 2^{n}$ , then $5 = 2^{2 m + n}$ which is impossible as 2 and 5 are prime so there are no positive integer solutions $m$ and $n$ of this equation.

(2) We have $a^{m} d^{n} = c^{n} b^{m}$ . As $a$ and $b$ are coprime, we get $a^{m} | c^{n}$ . Because $c$ and $d$ are coprime, so $c^{n} | a^{m}$ . This means $a^{m} = c^{n}$ . Similarly, $b^{m} = d^{n}$ .

If $a^{m} = c^{n}$ and $b^{m} = d^{n}$ , then obviously $(a / b)^{m} = (c / d)^{n}$ .

This implies that $a^{m}$ and $c^{n}$ have the same prime factors. Write $a = p_{1}^{u_{1}} . . . p_{k}^{u_{k}}$ and $c = p_{1}^{v_{1}} . . . p_{k}^{v_{k}}$ and for all $j$ we have ${mu}_{j} = {nv}_{j}$ so that

$\frac{u_{1}}{v_{1}} = \frac{u_{2}}{v_{2}} = . . . = \frac{u_{k}}{v_{k}} = \frac{m}{n} .$

Similarly for $b$ and $d$ . This is a very special necessary relationship between $a$ and $c$ and also between $b$ and $d$ so solutions rarely occur to the equation:

${(\frac{5}{4})}^{m} = {(\frac{2}{1})}^{n} .$

We now show this is a sufficient condition. Conversely suppose $a = p_{1}^{u_{1}} . . . p_{k}^{u_{k}}$ and $c = p_{1}^{v_{1}} . . . p_{k}^{v_{k}}$ and

$\frac{u_{1}}{v_{1}} = \frac{u_{2}}{v_{2}} = . . . = \frac{u_{k}}{v_{k}} .$

We call this common ratio $\frac{n}{m}$ then $u_{j} m = v_{j} n$ for all $j$ and $a^{m} = b^{n}$ . Similarly if corresponding ratios of the powers of the prime factors of $b$ and $d$ are constant and also equal to $\frac{n}{m}$ then $b^{m} = d^{n}$ giving

${(\frac{a}{b})}^{m} = {(\frac{c}{d})}^{n} .$