smallest n - application of Chinese remainder Theorem

Question

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?

Answer 1

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.

Answer 2

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 $\min(n)=2^{15}3^{10}5^6$