If there are integers

m

and

n

such that

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

then

\frac{\log 2}{\log 5 / 4}

is rational and

5^{m} = 2^{n + 2 m}

but this is impossible as 2 and 5 are prime so there cannot be integer solutions of this equation.

Now suppose

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

then $a^{m} d^{n} = b^{m} c^{n}$ and, as $a$ and $b$ are coprime, $a^{m}$ divides $c^{n}$ and $c^{n}$ divides $a^{m}$ so $a^{m} = c^{n}$ . Similarly $b^{m} = d^{n}$ .

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

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

Similarly for $b$ and $d$ .

This is a very special relationship between $a$ and $c$ and also between $b$ and $d$ so in general this does not happen and there are no positive integers $m, n$ such that

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

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

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