Why does "$p$ must occur in the prime factorization of $n^2 p$ either once or an odd number of times" $\implies$ "$p$ occurs as a factor of $m^2$ either once or an odd number of times"? --GFauxPas 23:45, 25 April 2012 (EDT)
- Because $n^2 p = m^2$. --abcxyz 01:19, 26 April 2012 (EDT)