Prove that $\sqrt {n − 1} +\sqrt {n + 1}$ is irrational for every integer $n ≥ 1$ This question is from Apostol's Mathematical Analysis.
I have a solution for it at SOLUTION!
The question is:
Prove that $\sqrt {n − 1} +\sqrt {n + 1}$ is irrational for every integer n ≥ 1.
And the proof I have is
Now, I can't understand how to get h=$3/2$ and k=$1/2$ ?
Please help.
 A: A different proof: Assume the expression is rational. Square it, to get $(\sqrt{n-1}+\sqrt{n+1})^2=2n+2\sqrt{n^2-1}$. This is rational, so $\sqrt{n^2-1}$ is rational, so it is an integer $m$, or $n^2-1=m^2$, which means that $n=1$ and $m=0$, but if $n=1$, then the original expression is $\sqrt2$, which is irrational, and we are done.
A: $(h+k)(h-k)=2$
$h+k$ and $h-k$ are integers and $h>k$
$\Rightarrow h+k=2$ and $h-k=1$
Solving, we get $h=3/2$ and $k=1/2$,  which is a contradiction.
A: We know that $k^2$ and $h^2$ are two positive integer squares that differ by $2$.  We could just stop here, since there are no such positive integers.
If this weren't immediately apparent, let $k \geq 1$ and $i \geq 1$, then $$(k+i)^2 = k^2+2ki+i^2 \geq k^2+3.$$  So two square integers differ by $3$ or more.
A: The following explains the mystery:-
Let start from “$n – 1 = k^2$ and $n + 1 = h^2$, where k and h are positive integer”.
Clearly, h > k and therefore we can let h = k + j for some positive integer j.
Then, $h^2 – k^2 = (n + 1) – (n – 1) = 2$
i.e. (h + k)(h – k) = 2
i.e. (2k + j)(j) = 2
This is a Diophantine type of equation saying that
“the product of two positive integers is 2, what can you say about these two numbers?”
Case 1 : (2k + j) = 1 and j = 2
--Solving them gives an unacceptable result of k = -1/2 (violating the assumption of k is positive).
Case 2 : (2k + j) = 2 and j = 1
-- Solving them gives k = 1/2
-- If k = 1/2, then h = k + j = 1/2 + 1 = 3/2
“which is (also) absurd.”
