prove that if a+ b/a - 1/b is an integer then it is a perfect square for integers a, b.
By setting a=-1, b=1 you can see the statement does not hold.
But let's try to fix it: "if a+ b/a - 1/b is an integer then it is a perfect square for positive integers a, b."
So let $\displaystyle a+\frac{b}{a}-\frac{1}{b} = a+\frac{b^2-a}{ab}$ be an integer for positive integers $\displaystyle a,b$. Then $\displaystyle b^2-a=kab$ for some integer $\displaystyle k$. We get $\displaystyle b^2=a(kb+1)$.
We see this implies $\displaystyle k \ge 0$.
If $\displaystyle k=0$ we immediately see that our statement holds.
We'll finish the proof by showing that we cannot have $\displaystyle k>0$. If it is, then $\displaystyle b^2=a(kb+1)$ implies $\displaystyle a<b$. After dividing $\displaystyle b^2=a(kb+1)$ by $\displaystyle b$ we get $\displaystyle b=ak+\frac{a}{b}$. This means that $\displaystyle b$ divides $\displaystyle a$, so we cannot have $\displaystyle a<b$, which is a contradiction.