Copyright © University of Cambridge. All rights reserved.
'Rarity' printed from https://nrich.maths.org/
Ping sent this solution from
Thailand.
(1) If $5^m=4^m 2^n$, then $5 = 2^{2m + n}$ 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 $${u_1\over v_1} =
{u_2\over v_2} = ... = {u_k\over v_k} = {m\over n}.$$ 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: $$\left({5\over 4}\right)^m = \left({2\over
1}\right)^n.$$ 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 $${u_1\over v_1} = {u_2\over v_2} = ...
= {u_k\over v_k}.$$ We call this common ratio ${n\over m}$ then
$u_jm=v_jn$ 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 ${n\over m}$ then $b^m=d^n$ giving
$$\left({a\over b}\right)^m = \left({c\over d}\right)^n.$$