4 ms·
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+k
by jerome-jh 7y ago
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 ;)
- boyobo 7y agoNot really sure what your argument is. But it’s wrong, because n=4 is a counterexample. The factors are 1,2 and 4. This is an odd number of factors. 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.
- jerome-jh 7y agoIndeed I was talking about prime factors. I understood your point later last night and it is certainly valid. I quite do not understand the original paper though.