Question

let n be a positive integer. If n has odd number of divisors(other than 1 and n) then show that n is a perfect square

Answer 1

let \( p_1^{a_1}p_2^{a_2}... p_r^{a_r}\) be the prime factorization of \(n\), then number of divisors of \(n\) is given by :

\((a_1 + 1)(a_2+1)...(a_r+1)\)

Answer 2

Notice that this expression is odd only if each of \(a_i\) is even. QED