[Hemanth_2u]

6.here are two whole numbers, difference of their squares is a cube and the difference of their cubes is a square. These are the smallest possible numbers. Can you find the numbers?




Ans:


Let the two positive numbers be a and b, with a > b. Now,
a^2 - b^2 = m^3 ==> (a+b)*(a-b) = m^3

It implies that (a-b) = m and (a-b)^2 = m^2. a = (m^2 + m)/2 and b =
(m^2 - m)/2. Now, there would be an infinite series satisfying this
condition that follows:

m a b
------------------
1 1 0
2 3 1
3 6 3
4 10 6
5 15 10
6 21 15
7 28 21
8 36 28
9 45 36
10 55 45
and, so on.

But, a^3 - b^3 = n^2 => n^2 = (a-b)*(a^2 + ab + b^2)
= (m^2/4)*(3*m^3 + m)

So, {3*(m^3) + m} should be a square as (m^2/4) is of that form. I am
completely stuck at this point. Please, help me out.

I could find out the following by just trial and error. The lowest
possible solution is (a,b) = (10,6)

10^2 - 6^2 = 64 = 4^3
10^3 - 6^3 = 784 = 28^2

But, I am nowhere near to finding a generic solution.

Eagerly waiting for your reply.

Thanks and regards