Let P = x^k, Q = x^m, k < m
Per my previous post, it follows that
x^(m-k) = (2m-k)/(2k-m)
where x, m and k are positive integers and k < m < 2k
Let a = m - k
Then
x^a = (2a + k)/(k - a) = 1 + 3a/(k-a)
where x, a and k are positive integers and a < k and x > 1
If a = 1, then
x = 1 + 3/(k-1) has two integral solutions:
a) k = 2, x = 4, m = k+a = k+1 = 3, P = x^k = 16, Q = x^m = 64
b) k = 4, x = 2, m = k+a = k+1 = 5, P = x^k = 16, Q = x^m = 32
If a = 2, then
x^2 = 1 + 6/(k-2) has one integral solution:
c) k = 4, x = 2, m = k+a = k+2 = 6, P = x^k = 16, Q = x^m = 64
If a = 3, then
x^3 = 1 + 9/(k-3) has no integral solutions.
the rhs is integral only if k = 4, 6, or 12
but then it equals 10, 4, or 2, none of which are cubes
And there are no integral solutions if a > 4, because
1 + 3a/(k-a) <= 3a + 1
and x^a >= 2^a > 3a
so x^a <> 1 + 3a/(k-a)
So the only solutions where P < Q are (P,Q) = (16,32) and (16,64)
And of course, the other solutions to the problem are (P,Q) = (32,16) and (64,16)
Edited on September 4, 2009, 2:08 pm
analytical solution, completed (spoiler)
