Solve $\sqrt{x-5}-\sqrt{9-x}\gt1,x\in\mathbb Z$ 
Solve $\sqrt{x-5}-\sqrt{9-x}\gt1,x\in\mathbb Z$

The statement tells us that $x\in[5,9]$. Also,
$$\sqrt{x-5}\gt1+\sqrt{9-x}$$
Since both sides are positive, we can square
$$x-5>1+9-x+2\sqrt{9-x}\\2x-15\gt2\sqrt{9-x}$$
$\implies 2x-15\gt0\implies x\gt7.5$
Since $x\in\mathbb Z\implies x=8,9$
But on back substitution, $x=8$ doesn't satisfy. Is there a way we could get the final answer without back sustitution?

Hint:
Let $\sqrt{x-5}=a\ge0$ and $\sqrt{9-x}=b\ge0$
$$\implies a^2+b^2=4$$
and $a-b>1\iff a> b+1$
$$4=a^2+b^2>(b+1)^2+b^2\iff 2b^2+2b-3<0$$
Now for $(x-a)(x-b)<0, a<b;$ $$a<x<b$$
