Find all pairs of positive integers x and y such that 2x+1 is divisible by y and at the same time 2y+1 is divisible by x.
Other solutions and proofs that I have seen did not actually give all the possible answers or explain that the solutions they have given are the ONLY possible answers.
Assistance would be greatly appreciated, as would a comprehensive and detailed proof outlining everything clearly.
Since `2x + 1` is divisible by y, hence, dividing `2x+1` by y yields a reminder equal to 0 such that:
`(2x+1)/y = k => ky = 2x + 1 => 2x = ky - 1 => x = (ky-1)/2`
k is integer
Since x is an integer, hence `(ky-1)` is a multiple of 2.
Since `2y + 1` is divisible by x, hence, dividing `2y+1` by x yields a reminder equal to 0 such that:
`(2y+1)/x = p => px = 2y + 1 => 2y = px - 1 => y = (px - 1)/2`
p is integer
`y = (p(ky-1)/2 - 1)/2 => y = (pky - p - 2)/4 => 4y - pky = -p - 2`
Factoring out y yields:
`y(pk - 4) = p + 2 => y = (p + 2)/(pk - 4)`
Substituting `(p + 2)/(pk - 4)` for y in `x = (ky-1)/2` yields:
`x = (k(p + 2)/(pk - 4) - 1)/2 => x = (pk + 2k - pk + 4)/(2(pk-4))`
`x = 2(k+2)/(2(pk-4)) => x = (k+2)/(pk-4)`
Hence, evaluating the integers x and y, under the given conditions, yields `x = (k+2)/(pk-4)` and `y = (p + 2)/(pk - 4), ` where `(p + 2)` is a divisor of `k+2` and `p+2.`