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

4,897

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.

Share:
4,897

Related videos on Youtube

Aks
Author by

Aks

Updated on September 12, 2022

Comments

  • Aks
    Aks about 1 year

    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 $n_0$.

  • JiK
    JiK about 9 years
    I'm assuming you have defined $n=ym$ where $m$ is not divisible by $2$, $3$ or $5$?
  • Michael Stocker
    Michael Stocker about 9 years
    Aks changed the variable names around a few times. y was at one time the smallest possible solution. Any additional primes factor would obviously make the solution bigger.
  • Aks
    Aks about 9 years
    does this mean that a=b=c=1?since n>0
  • Michael Stocker
    Michael Stocker about 9 years
    No. You have to find the smallest solution to all the congruences. The second congruence is $b≡ 0mod 2$, so b must be even and can't be 1. Take a for example. We have from $a-1≡0mod2$, $a≡0mod3$, $a≡0mod5$ that a is odd, and divisible by 3 and 5. The smallest a possible is therefore 15.