75th NMO Selection Tests

Question

We say that a natural number n 3 is almost square-free if there exists a prime number p, with p 1 3, such that n is divisible by p^2, and the number np is square-free (i.e., not divisible by the square of any prime number). Show that, for any natural number n that is almost square-free, the ratio between twice the sum of the divisors of n and the number of divisors of n is a natural number. Lucian Petrescu

Accepted Answer

Consider n = p_1 p_k-1 p_k^2 p_k+1 p_s, with p_1 < p_2 < < p_s prime numbers, 1 k s, and p = p_k 1 3. Moreover, p is odd and satisfies p^2 + p + 1 0 3. The number of divisors of n, (n), is (n) = (1+1) (1+1) (1+1)_applied s-1 times (2+1) = 3 2^s-1, and the sum of the divisors, (n), is: aligned (n) &= p_1^2-1p_1-1 p_k-1^2-1p_k-1-1 p_k^3-1p_k-1 p_k+1^2-1p_k+1-1 p_s^2-1p_s-1 &= (p_1+1) (p_k-1+1) (p_k^2+p_k+1) (p_k+1+1) (p_s+1). aligned **Case I.** If n is even, then p_1 = 2, and p_2 < p_3 < < p_s are odd primes. Hence, 2(n) = 2 (2+1) (p_2+1) (p_k-1+1) (p_k^2+p_k+1) (p_k+1+1) (p_s+1), so 2(n) is divisible by 2 3^2 2^s-2 = 3 (n). **Case II.** If n is odd, then p_1 < p_2 < < p_s are odd primes, so (n) = (p_1 + 1) (p_k-1 + 1) (p_k^2 + p_k + 1) (p_k+1 + 1) (p_s + 1) is divisible by 3 2^s-1 = (n).