Find the smallest numbers $a,b$, and $c$ such that: $a^2=2b^3=3c^5.$ What can you say about other solutions to this problem? Congratulations for your good solutions to Ella and Elizabeth , S6, Madras College and Yiwan, The Chinese High Singapore. Here is Yiwan's solution:

${\displaystyle a^2=2b^3=3c^5.}$

As (2,3)=1, that is 2 and 3 have no common divisor other than 1, we shall write $a$, $b$, and $c$ in terms of powers of 2 and 3. Let $c=2^p{3}^q$ (where p, q are integer numbers above 0). Then

${\displaystyle 3c^5=2^{5p}{3}^{5q+1}=2b^3=a^2.}$

Hence

${\displaystyle b=2^{(5p-1)/3}{3}^{(5q+1)/3};}$

${\displaystyle a=2^{5p/2}{3}^{(5q+1)/2}.}$

As a, b are all integers, it follows that $3|(5p-1)$, $3|(5q+1)$, $2|(5p)$ and $2|(5q+1)$ [using the notation $3|(5p-1)$ to mean 3 divides or is a factor of $(5p-1)$]. Obviously the solution for the smallest number is when p=2 and q=1. In this case, $c=2^2\times 3=12$ ; $b=2^3\times 3^2=72$ ; $a=2^5\times 3^3=864.$ The smallest solution is $(a=864,b=72,c=12).$ For other solutions take

${\displaystyle p=\{2,8,14,20,...6m+2\}}$

where m is a positive integer and

${\displaystyle q=\{1,7,13,19,...6n-5\}}$

where n is a positive integer. If we substitute any value of $p$ and $q$ from the corresponding domain, we will get the other solutions for the equation.