If x and y are odd integers that satisfy the equation:
x*y2+x = 2*x2018+y2+3
Determine all possible values of logxy.
I will note that if x is negative then the left side of the equation is negative while the right side is strictly positive. Thus there are no solutions with negative x.
y only occurs as a square, so any positive has a matching negative solution, but the requested calculation of log_x(y) goes into multivalued complex domain for either x or y negative in what is otherwise a high-school level challenge problem. So I will discard the negative option for y as well.
Now, rearrange the equation to isolate y^2, and a few more bits of manipulation:
(x-1)*y^2 = [2x*(x^2017-1)] - [x-1] + 4
The left side has x-1 as a factor and the two terms in brackets on the right side also have x-1 as a factor. Therefore x-1 must divide 4.
Positive odd integers x which satisfy this are x=3 and x=5.
If x=3 then 3y^2 + 3 = 2*3^2018 + y^2 + 3, which simplifies to y^2 = 3^2018. Then y=3^1009.
Then log_x(y) = log_3(3^1009) = 1009.
If x=5 then 5y^2 + 5 = 2*5^2018 + y^2 + 3, which simplifies to 2y^2 = 5^2018 - 1. In this case y is not an integer.
I will note that 2018 can be replaced with any positive even integer, call it 2N. Then the equation will have at least one solution where log_x(y) = N.
Solution