We will prove that, searching for the rational solutions.

Supposing that #sqrt(n+1)+sqrt(n-1)# is rational then

#sqrt(n+1)+sqrt(n-1)=p_1/q_1# so

#(sqrt(n+1)-sqrt(n-1))(sqrt(n+1)+sqrt(n-1))=n+1-(n-1)=2 = (sqrt(n+1)-sqrt(n-1))p_1/q_2# is also rational

So #sqrt(n+1)-sqrt(n-1) = p_2/q_2# is also rational then

#2sqrt(n+1) = p_1/q_1+p_2/q_2 = (p_1q_1+p_2q_2)/(q_1q_2)#

So #sqrt(n+1) = 1/2 (p_1q_1+p_2q_2)/(q_1q_2)#

With the same argument we could prove also that

#sqrt(n-1)# is rational.

So #sqrt(n^2-1) = p/q# is rational and consequently

#n^2 = (p/q)^2+1# Supposing that #p/q = m# ( remember that #n# is integer) we will need

#n^2=m^2+1# or

#(n+m)(n-m) =1# with #n,m# integers, #m = 0, n= 1# being the only rational solution.

Concluding, the only rational solution is for #m = 0, n=1# but then

#sqrt2# is irrational so for #n in NN^+#, #sqrt(n+1)+sqrt(n-1)# is irrational.