Exercise 1.3.16 ( $n/2$ is a square, $n/3$ is a cube and $n/5$ is a fifth power)

Question

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

Richard Ganaye · Accepted Answer

\begin{proof} We want $ n = 2 a^2 = 3b^3 = 5 c^5$. Try
\begin{align*}
a &= 2^r 3^s 5^t,\
b &= 2^u 3^v 5^w,\
c &= 2^x 3^y 5^z,
\end{align*}
where $r,s,t,u,v,w,x,y,z$ are unknown. Then
\begin{align*}
2 a^2 = 3b^3 = 5 c^5  \iff &2^{2r+1} 3^{2s} 5^{2t} = 2^{3u}3^{3v+1} 5^{3w} = 2^{5x}3^{5y} 5^{5z+1}\
\iff  &
\left\{
\begin{array}{l}
 2r+1 = 3u = 5x,\
2s = 3v + 1 = 5 y,\
2t = 3w = 5z + 1.
\end{array}
ight.
\end{align*}
Note that $5 \mid u , 3 \mid x, 5 \mid s, 2 \mid y, 3 \mid t, 2 \mid w$. This help us to find a solution
$$(r,u,x) = (7,5,3), \qquad (s,v,y) = (5,3,2),\qquad (t,w,z) = (3,2,1).$$
This gives a solution, among many others,
$$n = 30233088000000 = 2^{15} \cdot  3^{10}  \cdot 5^6 = 2(2^7 \cdot 3^5 \cdot 5^3)^2 = 3( 2^5\cdot 3^3 \cdot 5^2)^3 = 5(2^3 \cdot 3^2 \cdot 5)^5.$$
\end{proof}

Exercise 1.3.16 ( $n/2$ is a square, $n/3$ is a cube and $n/5$ is a fifth power)

Answers

Comments

Exercise 1.3.16 ( $n/2$ is a square, $n/3$ is a cube and $n/5$ is a fifth power)

Answers

Comments

Add answer