Here's what I have so far
Without loss of generality, assume that P < Q.
P and Q must both be powers of a common factor, x.
Let P = x^k, Q = x^m, k < m
Substituting gives
(x^k)^(x^k +2x^m) = (x^m)^(x^m + 2x^k)
Associately,
x^(kx^k +2kx^m) = x^(mx^m + 2mx^k)
So kx^k +2kx^m = mx^m + 2mx^k
Rearranging gives
x^(m-k) = (2m-k)/(2k-m)
Because the lhs and the numerator are positive, it follows that
the denominator (2k - m) is also positive, so
k < m < 2k
But that's as far as I got.
Can anybody solve
x^(m-k) = (2m-k)/(2k-m)
where x, m and k are positive integers and k < m < 2k?
Or how about letting a = m - k?
Then
x^a = (2a + k)/(k - a)
where x, a and k are positive integers and a < k ?
analytical attempt
