smallest n - application of Chinese remainder Theorem I am trying to find the smallest $n \in \mathbb{N}\setminus \{ 0 \}$, such that $n = 2 x^2 = 3y^3 = 5 z^5$, for $x,y,z \in \mathbb{Z}$. Is there a way to prove this by the Chinese Remainder Theorem?
 A: As @MatthewDaly said it must be of form $n=2^a3^b5^c$. Although it can also has different primes, but if you want the smallest one, you should avoid to include other non necessary ones. Now suppose:
$$x=2^{a'}3^{b'}5^{c'}$$
$$y=2^{a''}3^{b''}5^{c''}$$
$$x=2^{a'''}3^{b'''}5^{c'''}$$
So we have:
$$a=2a'+1=3a''=5a''' \Longrightarrow a=30a_0+15$$
$$b=2b'=3b''+1=5b''' \Longrightarrow b=30b_0+10$$
$$c=2c'=3c''=5c'''+1 \Longrightarrow c=30a_0+6$$
So the smallest integer is $\min(n)=2^{15}3^{10}5^6$
A: The question asks to find the smallest positive integer that is simultaneously twice a square, three times a cube, and five times a fifth power.
Writing $n$ as $2^a \cdot 3^b \cdot 5^c$ and assuming that no other prime divides $n$, it follows that the following congruences must hold:


*

*$a$ is:


*

*$\equiv 1 \pmod{2}$

*$\equiv 0 \pmod{3}$

*$\equiv 0 \pmod{5}$


*$b$ is:


*

*$\equiv 0 \pmod{2}$

*$\equiv 1 \pmod{3}$

*$\equiv 0 \pmod{5}$


*$c$ is:


*

*$\equiv 0 \pmod{2}$

*$\equiv 0 \pmod{3}$

*$\equiv 1 \pmod{5}$
The smallest values for $a$, $b$, and $c$ (which exist by the Chinese remainder theorem) are thus $15$, $10$, and $6$ respectively, so the smallest solution is $n=2^{15} \cdot 3^{10} \cdot 5^6=30,233,088,000,000$.
By the way, it happens that each prime factor of $30$ divides one less than the product of the other prime factors. A squarefree composite number with this property is called a Giuga number.
