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## 'Em'power'ed' printed from http://nrich.maths.org/

Find the smallest numbers $a, b$, and $c$ such that: $a^2 = 2 b^3 =
3 c^5.$ 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: $$a^2=2b^3=3c^5.$$ 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 $c=2^p3^q$ (where p, q are
integer numbers above 0). Then $$3c^5=2^{5p}3^{5q+1}=2b^3=a^2.$$
Hence $$b=2^{(5p-1)/3}3^{(5q+1)/3};$$ $$a=2^{5p/2}3^{(5q+1)/2}.$$
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 $$p=\{2, 8, 14, 20, ... 6m + 2\}$$ where m is a positive
integer and $$q=\{1, 7, 13, 19, ... 6n - 5\}$$ 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.