Find all positive integers n such that the number of positive divisors of n is (4n)1/3.
Solution
Answer: 2, 128, 2000.
Let e(p) be the exponent of p in the prime factorisation of n (so n = 2e(2)3e(3)5e(5) ... ). Since 4n is a cube, we must have e(2) = 1 mod 3 and e(p) = 0 mod 3 for odd p. Let d(n) be the number of positive divisors of n. Then d(n) = (1 + e(2) )(1 + e(3) )(1 + e(5) ) ... . Now 1 + e(2) = 2 mod 3 and 1 + e(p) = 1 mod 3 for p > 2, so d(n) is not a multiple of 3. But 4n = d(n)3, so n is not a multiple of 3, so e(3) = 0.
Put e(2) = 3k + 1, and e(p) = 3 f(p) for odd primes p. Then d(n) = (4n)1/3 gives (3k + 2) ∏p>3 (1 + 3 f(p) ) = 2k+1 ∏p>3 pf(p). So (3k + 2)/2k+1 = ∏p>3 pf(p)/(1 + 3 f(p) ) (*).
Now for p ≥ 5, we have pf(p) ≥ (1 + 4)f(p) ≥ 1 + 4 f(p) > 1 + 3 f(p), unless f(p) = 0. So the rhs of (*) is > 1 if any odd prime > 5 divides n and = 1 if not. But 2k+1 > 2 + 3k for k > 2 (eg a trivial induction). So the lhs of (*) is < 1 for k > 2. So we must have k = 0, 1 or 2.
If k = 0 or 2, then 2k+1 = 2 + 3k, so the rhs of (*) is 1, so n has no prime factors except 2. That gives the solutions n = 2, 128. If k = 1, then (3k + 2)/2k+1 = 5/4. Now if p >= 7, then ph ≥ (1 + 6)h ≥ 1 + 6h. So if n has a prime factor greater than 5, then the rhs of (*) is at least (1 + 6h)/(1 + 3h) with h ≥ 1, and hence at least 3/2 ( h ≥ 1 implies 3/2h > 1/2 implies 1 + 6h > 3/2 (1 + 3h) ). So n cannot have a prime factor greater than 5. Suppose f(5) = h. Then (*) gives 5/4 = 5h/(1 + 3h), which implies h = 1 (5h/(1 + 3h) is a strictly increasing function of h, so h = 1 is the only solution). Thus we get n = 2453 = 2000.
© John Scholes
jscholes@kalva.demon.co.uk
10 Oct 2002