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?

There is no maximum value of $y$:
Let $y$ be any integer with $y\ge3$, and let $x=y-1$.
Then $x^2-y^2<0<8$, and $x+y=2y-1\ge5>3$.
