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

2 Answers
Aug 24, 2016

Answer:

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

Answer:

#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