Question

For what integer values of `k` does `sqrt(x)+sqrt(x+1)=k` have a rational solution?

Answer 1

Any positive integer `k` gives a rational solution.

Explanation:

Given any positive integer `k`, let `x = (k^2-1)^2/(4k^2)`

Note that this value of `x` is rational.

Then:

`sqrt(x) + sqrt(x+1)`

`=sqrt((k^2-1)^2/(4k^2))+sqrt((k^2-1)^2/(4k^2)+1)`

`=sqrt((k^2-1)^2/(4k^2))+sqrt(((k^2-1)^2+4k^2)/(4k^2))`

`=sqrt((k^2-1)^2/(4k^2))+sqrt(((k^2+1)^2)/(4k^2))`

`=(k^2-1)/(2k)+(k^2+1)/(2k)`

`=k^2/k`

`=k`

Answer 2

`x = ( (k^2-1)/(2k))^2` for `k in NN-{0}` are the rational solutions

Explanation:

`sqrt(x)+sqrt(x+1)=k`. If `x` is rational, and `k` is an integer then

`x = (n/m)^2` so

`n/m + sqrt((n/m)^2+1) = k` and

`sqrt((n/m)^2+1) = k-n/m`

so

`(n/m)^2+1 =(k-n/m)^2` or

`1 = k^2-2kn/m` then

`n/m = (k^2-1)/(2k)`

Finally

`x = ( (k^2-1)/(2k))^2` for `k in NN-{0}` are the rational solutions