A fun question.
$\displaystyle 4^3 = 8^2$
Are there any more pairs of numbers (a, b) such that $\displaystyle a^3 = b^2$ - if so how many?
I would be interested in how people go about finding the solution .
A fun question.
$\displaystyle 4^3 = 8^2$
Are there any more pairs of numbers (a, b) such that $\displaystyle a^3 = b^2$ - if so how many?
I would be interested in how people go about finding the solution .
There's an infinite number of such pairs. Letting $\displaystyle (a,b) = (n^2,n^3)$ works, where $\displaystyle n$ is an integer. In fact, $\displaystyle a$ has to be a square.
Why? Consider prime factorizations of $\displaystyle a$ and $\displaystyle b$, suppose $\displaystyle a^3=b^2$, and use uniqueness of prime factorizations:
$\displaystyle a = p_1^{x_1}\cdots p_k^{x_k},$
$\displaystyle b = p_1^{y_1}\cdots p_k^{y_k},$
where the equation $\displaystyle a^3=b^2$ implies that
$\displaystyle p_i^{3x_i} = p_i^{2y_i}$
for all $\displaystyle i=1,2,\ldots,k$. Hence $\displaystyle 3x_i = 2y_i$, whence $\displaystyle x_i$ must be even. In conclusion, each $\displaystyle p_i^{x_i}$ is a square, and so all of $\displaystyle a$ is square.
I answered this by generating number by raising a base number to the powers 2 and 3.
$\displaystyle a = n^2$
$\displaystyle b = n^3$
substitute into the expression
$\displaystyle a^3 = b^2$
gives
$\displaystyle (n^2)^3 = (n^3)^2 {\Rightarrow} n^6 = n^6$
Example
$\displaystyle n = 3, n^2 = 9, n^3 = 27$
so does $\displaystyle 9^3 = 27^2$ (yes it does)
There are infinate solutions because $\displaystyle n {\in} {\Re}$