text: Q:
  
  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:
  
  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:
  
  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.
  
  