1

How can I mathematically prove that the squares of two relatively prime numbers are also relatively prime?

Thanks in advance.

Peter Ye
  • 160
  • 1
    Ye it is a duplicate but really there's a much easier way to solve this. Just factor the numbers into product of primes and then square. – Yanko Dec 08 '17 at 17:26
  • 1
    I disagree that it's a duplicate. The linked problem is asking for help with a specific proof method. – Sort of Damocles Dec 08 '17 at 17:36

1 Answers1

2

Suppose $x = 2^{m_2}\cdot 3^{m_3} \cdots \cdot p_r^{m_r}$ and $y = 2^{n_2}\cdot 3^{n_3}\cdots p_s^{n_s}$. If their gcd is 1, then whenever $m_i>0$ then $n_i=0$ and vice-versa (you can take this as the definition of $gcd(x,y)=1$).

Now $x^2 = 2^{2m_2}\cdot 3^{2m_3} \cdots \cdot p_r^{2m_r}$ and $y^2 = 2^{2n_2}\cdot 3^{2n_3}\cdots p_s^{2n_s}$. It is clear that $n=0$ iff $2n=0$, and $n>0$ iff $2n>0$, so the squares must also be relatively prime.

Sort of Damocles
  • 2,443