Determine all possible values of a positive integer P, such that P-1 has three divisors Q, R and S, with Q < R < S, such that: Q + R + S = P

Divide Q + R + S = P by (P-1) to get

Q/(P-1) + R/(P-1) + S/(P-1) = P/(P-1) = 1 + 1/(P-1)

So we need 3 unit fractions that sum to one more that a unit fraction call them
1/x + 1/y + 1/z = 1 + 1/(P-1)
where 1/x < 1/y < 1/z

If z=1 we have
1/x + 1/y = 1/(P-1)
(x+y)/(xy) = 1/(P-1)
x = ((P-1)y)/(y-(P-1))
so
(y-(p-1))/((p-1)y) + 1/y = 1 = 1/(P-1)
P-1 = 1
P=2
which is impossible

If z=2
We need the other 2 fractions to sum to more than 1/2
1/4 + 1/3 + 1/2 = 13/12
1/5 + 1/3 + 1/2 = 31/30
are the only ways

If z > 2 we will not have solutions because the largest fraction has to be greater than 1/3

So the solutions are P=13 = 3+4+6 and P=31 = 6+10+15

Posted by Jer on 2010-12-17 16:26:14