6th Balkan 1989 Problem 1 Find all integers which are the sum of the squares of their four smallest positive divisors. Solution
: 130. n cannot be odd, because then its four smallest divisors would be odd and hence the sum of their squares even. So the two smallest divisors are 1 and 2. If 4 divides n, then the four smallest divisors are either 1, 2, 4, 8 or 1, 2, 4, p (where p may be 3 or 5 or more). In the first case the sum of the squares is odd, but n is even, so it fails. In the second case we have 21 + p 2 = 4pm, where m is the product of the other factors of n. But any solution must divide 21, so p must be 3 or 7. But then 21 + p 2 is not divisible by 4, so this does not give any solutions. So we can assume 2 divides n, but not 4. So if p is the smallest odd prime dividing n, then the three smallest divisors are 1, 2 and p. If the next smallest is q, then three of the four smallest are odd and hence the sum of the squares is odd. So the next smallest must be 2p. Thus we have 5 + 5p 2 = 2pm. So 5 divides n, so p must be 3 or 5. p = 3 gives square sum 50, which fails (not divisible by 3). p = 5 gives n = 130, which work. 6th Balkan 1989 © John Scholes jscholes@kalva.demon.co.uk 11 Apr 2002