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 Unequally Yoked III (Posted on 2009-09-01)
Determine all possible nonzero integer(s) P and Q, with PQ, that satisfy this equation.

PP+2Q = QQ+2P

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes)

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 analytical attempt | Comment 2 of 3 |
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 ?

 Posted by Steve Herman on 2009-09-02 14:35:16

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