N is the product of three primes, p,q,r. Then the sum of the divisors of N is 1+p+q+r+p*q+p*r+q*r+p*q*r = (1+p)(1+q)(1+r) = 10560

10560 has 11 as a factor, so testing multiples of 11 looking for a number one greater than a prime gives 4*11-1=43, 10*11-1=109, 12*11-1=131, etc. The first two are too large for the sum of squares to equal 2331, so 43 is one of the factors.

Then, 10560/44 = 240 = (1+q)(1+r) and 2331-43^2 = 482 = q^2+r^2. Which makes q and r equal 11 and 19.

N = 43*19*11 = **8987**.