Solution of inequality Question:
$x$ and $y$ are different integers and both $x$ and $y$ are greater than $0$. If  $x^2-y^2<8$ and $x+y>3$, what is the maximum value of $x$?
My solution:
I used plugged in number method, and found for $x=4$ and $y=3$, $4^2-3^2=7$ and $4+3=7$, satisfies both equations. but, if $x=5$ and $y=4$, values doesn't satisfy first equation. hence, maximum possible value of $x= 4$.
Is there any other easy way than this to solve this problem?

As $x^2 - y^2 = (x-y)(x+y) < 8 $, and $x+y > 3$ we know $x-y<2$.
So, if there is a additional condititon  $x \neq y$ (I will asume that, also that $x>y$), then $x=1+y$.
So $(y+1)^2 - y^2 < 8$, so $2y+1<8 $, and maximum of $y$ is $3$.
Without the condition $x \neq y$ the value of $y$ is unbounded. 
