perfect square => even number of factors
n = r^2
n = (p1^k1 * p2^k2 * ... * pi^ki)^2
n = p1^(2 * k1) * p2^(2 * k2) * ... * pi^(2 * ki)
hence n has 2 * (k1+k2+...+ki) factors. Or am I wrong?
Reciprocal left as an exercise ;)
n = r^2
n = (p1^k1 * p2^k2 * ... * pi^ki)^2
n = p1^(2 * k1) * p2^(2 * k2) * ... * pi^(2 * ki)
hence n has 2 * (k1+k2+...+ki) factors. Or am I wrong?
Reciprocal left as an exercise ;)
Edit: it seems you are talking about 'The number of factors in the prime factorization of n'. This is not really relevant to the question posed in the submission.