Problem:
Let Ï„(n) denote the number of positive integer divisors of n. Find the sum of the six least positive integers n that are solutions to Ï„(n)+Ï„(n+1)=7.
Solution:
Let p,q, and r represent primes. Because τ(n)=1 only for n=1, there is no n for which {τ(n),τ(n+1)}={1,6}. If {τ(n),τ(n+1)}={2,5}, then {n,n+1}={p,q4}, so ∣∣∣​p−q4∣∣∣​=1. Checking q=2 and p=17 yields the solution n=16. If q>2, then q is odd, and p=q4±1 is even, so p cannot be prime.\
If {τ(n),τ(n+1)}={3,4}, then {n,n+1}={p2,q3} or {p2,qr}. Consider ∣∣∣​p2−q3∣∣∣​=1. If p2−1=(p−1)(p+1)=q3, then q=2. This yields the solution p=3 and q=2, so n=8. If q3−1=(q−1)(q2+q+1)=p2, then q−1=1, which does not give a solution. Consider ∣∣∣​p2−qr∣∣∣​=1. If p2−1=(p−1)(p+1)=qr, then if p>2, the left side is divisible by 8 , so there are no solutions. Finding the smallest four primes such that p2+1=qr gives 32+1=10,52+1=26,112+1=122, and 192+1=362. The six least values of n are 8,9,16,25,121, and 361 , whose sum is 540 .
The problems on this page are the property of the MAA's American Mathematics Competitions