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?

Starting from where you left off, since both sides of the inequality $$2x - 15 > 2\sqrt{9 - x}$$ are positive, the direction of the inequality is preserved if we square both sides, which yields
\begin{align*}
4x^2 - 60x + 225 & > 4(9 - x)\\
4x^2 - 60x + 225 & > 36 - 4x\\
4x^2 - 56x & > -189
\end{align*}
Since $(2a + b)^2 = 4a^2 + 4ab + b^2$, we can complete the square with $a = x$ and $b = 14$ to obtain
\begin{align*}
4x^2 - 56x + 196 & > 7\\
4(x^2 - 14x + 49) & > 7\\
(x - 7)^2 & > \frac{7}{4}
\end{align*}
Since $(8 - 7)^2 = 1 < \dfrac{7}{4}$, this eliminates $8$.  Thus, the only integer solution is $x = 9$.
