What is the smallest positive integer n such that 2n is a perfect square and 3n is a perfect cube and 5n is a perfect fifth power?

Guest Jun 17, 2021

#1**+1 **

From the assumptions it follows that 2, 3, and 5 are all divisors of $n$. Also $n$ should not be divisible by any other primes because if it were then dividing $n$ by the product of all these other primes would result in still another integer satisfying all the given assumptions.

Thus we conclude that $n= 2^a\cdot 3^b\cdot 5^c$ for some nonnegative integer $a$, $b$, and $c$.

The assumption $2n$ being a perfect square then implies $2 \mid a+1$, $2 \mid b$, and $2 \mid c$.

The assumption $3n$ being a perfect cube then implies $3 \mid a$, $3 \mid b+1$, and $3 \mid c$.

The assumption $5n$ being a perfect fifth power then implies $5 \mid a$, $5 \mid b$, and $5 \mid c+1$.

So $a$ has to be a multiple of ${\rm lcm}(3, 5)$ such that $a+1$ is even,

$b$ has to be a multiple of ${\rm lcm}(2, 5)$ such that $b+1$ is divisible by 3,

and $c$ has to be a multiple of ${\rm lcm}(2, 3)$ such that $c+1$ is divisible by 5.

The least numbers satisfying these conditions are $a=15$, $b=20$, and $c=24$.

Thus $n=2^{15}\cdot 3^{20}\cdot 5^{24}$.

Bginner Jun 18, 2021