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

2 Answers
Aug 24, 2016

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

Aug 24, 2016

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