MathLabs

Problem 3

For any positive integer n n , let d(n) d(n) denote the number of positive divisors of n n (including 11 and n n itself). Determine all positive integers k k such that d(n2)d(n)=k\frac{d(n^2)}{d(n)}=k for some n n .
Step 1 of 6: Express the divisor ratio
In plain words

Prime exponents control both divisor counts.

n=∏i=1rpiai⇒d(n2)d(n)=∏i=1r2ai+1ai+1n=\prod_{i=1}^r p_i^{a_i}\quad\Rightarrow\quad \frac{d(n^2)}{d(n)}=\prod_{i=1}^r\frac{2a_i+1}{a_i+1}
Detailed analysis

For the prime factorization n=∏piai n=\prod p_i^{a_i}, the divisor formula gives d(n)=∏(ai+1) d(n)=\prod(a_i+1) and d(n2)=∏(2ai+1) d(n^2)=\prod(2a_i+1). Hence the ratio is the displayed product. Each factor is rational, but the whole product is the integer k k in the problem.