Find the smallest numbers a, b, and c such that: a² = 2 b³ = 3 c⁵. 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:

a2=2b3=3c5.

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

3c5=25p35q+1=2b3=a2.

Hence

b=2(5p-1)/33(5q+1)/3;

a=25p/23(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=2² ×3=12 ; b=2³ ×3²=72 ; a=2⁵ ×3³=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.