Find $n$ such that $n/2$ is a square, $n/3$ is a cube, and $n/5$ a fifth power

Consider the set of positive integers $n \in \mathbb {Z}>0$ such that $\dfrac{n}{2}$ is a perfect square, $\dfrac{n}{3}$ is a perfect cube, and $\dfrac{n}{5}$ is a perfect fifth power; that is to say, such that there exist integers $a, b, c \in \mathbb{Z}$ such that $\dfrac{n}{2} = a^2$, $\dfrac{n}{3} = b^3$, $\dfrac{n}{5} = c^5$.

Let $n_0$ denote the smallest element in this set (it exists by the well-ordering principle). Determine the numerical value of $n_0$.

Solution 1:

Consider the prime factorization of n. n must be divisible by 2, 3 and 5. So y must be of the form

$y = 2^a · 3^b · 5^c$

and we get

$\frac{y}{2} = 2^{a-1} · 3^b · 5^c = d^2$

$\frac{y}{3} = 2^{a} · 3^{b-1} · 5^c = e^3$

$\frac{y}{5} = 2^{a} · 3^b · 5^{c-1} = f^5$

looking at the exponents we have

$a-1\equiv 0 \mod{2}$ , $b\equiv 0 \mod{2}$ , $c\equiv 0 \mod{2}$ $a\equiv 0 \mod{3}$ , $b-1\equiv 0 \mod{3}$ , $c\equiv 0 \mod{3}$ $a\equiv 0 \mod{5}$ , $b\equiv 0 \mod{5}$ , $c-1\equiv 0 \mod{5}$

Now you have to solve those equations for the smallest possible values of a,b and c and are done.