1. ## square = cube

A fun question.

$4^3 = 8^2$

Are there any more pairs of numbers (a, b) such that $a^3 = b^2$ - if so how many?

I would be interested in how people go about finding the solution .

2. There's an infinite number of such pairs. Letting $(a,b) = (n^2,n^3)$ works, where $n$ is an integer. In fact, $a$ has to be a square.

Why? Consider prime factorizations of $a$ and $b$, suppose $a^3=b^2$, and use uniqueness of prime factorizations:

$a = p_1^{x_1}\cdots p_k^{x_k},$
$b = p_1^{y_1}\cdots p_k^{y_k},$

where the equation $a^3=b^2$ implies that

$p_i^{3x_i} = p_i^{2y_i}$

for all $i=1,2,\ldots,k$. Hence $3x_i = 2y_i$, whence $x_i$ must be even. In conclusion, each $p_i^{x_i}$ is a square, and so all of $a$ is square.

3. I answered this by generating number by raising a base number to the powers 2 and 3.

$a = n^2$
$b = n^3$

substitute into the expression

$a^3 = b^2$

gives

$(n^2)^3 = (n^3)^2 {\Rightarrow} n^6 = n^6$

Example

$n = 3, n^2 = 9, n^3 = 27$

so does $9^3 = 27^2$ (yes it does)

There are infinate solutions because $n {\in} {\Re}$