$\forall x,y\in \mathbb N\ ,\ \exists\ z\in\mathbb N$ that $x+z$ is square but $y+z$ is not square. I am trying to prove this:


  $\forall x,y\in \mathbb N$ and $x\neq y,\  \exists\  z\in\mathbb N$ that $x+z$ is
  square but $y+z$ is not square.


$\mathbb N$ is set of natural numbers. 
Can you please suggest a hint?
You know how far apart $x$ and $y$ are.  Look for squares that are further apart than that.
Suppose the contrary; that is, suppose that for each pair $x$, $y$ of distinct natural numbers, the following is true: 

For every natural umber $z$, if $x+z = m^2$ for some natural number $m$, then $y+z = n^2$ for some (other) natural number $n$. 

Obviously, $m \neq n$ since $x\neq y$. 

Then we must have $$z = m^2 - x = n^2 - y,$$ 
and so we must have $$x-y = m^2 - n^2,$$ 
Which is clearly false, for example, when $x=2$ and $y=1$ for the smallest difference (in absolute value) between the squares of two natural numbers is $3$. In any case, there are no natural numbers $m$ and $n$ whose squares differ by $1$. So a contradiction arises, and our assertion is proved.