Find $x$ such that $\sqrt{x+\sqrt{x+7}}\in \mathbb{N}$ Find $x$ such that $$\sqrt{x+\sqrt{x+7}}\in \mathbb{N}$$
I tried many ways: $$\sqrt{x+\sqrt{x+7}}=n$$
$$\sqrt{x+\sqrt{x+7}}^2=n^2$$
$$x+\sqrt{x+7}=n^2$$
then solve for $x$ but didn't do with success.
I think this is the most difficult problem in my lifetime
Also $x$ must be made
 of $2$ digit.

Thanks to everybody for helping me understand this problem and its solution!


We want to find $a \in \mathbb{N}$ such that
$$a = \sqrt{x + \sqrt{x + 7}}.$$
Also implicit in the question was that $x \in \mathbb{N}$, but that wasn't stated, only implied by the statement that $x$ is two-digit.
Let $n = \sqrt{x + 7}$. Then $x = n^2 - 7$. Substituting into the original equation:
$$\begin{align}
a = & \sqrt{n^2 + n - 7}\\
0 = & n^2 + n - 7 - a^2
\end{align}$$
Solving for $n$:
$$n = -\frac{1}{2} \pm \frac{1}{2}\sqrt{4 a^2 + 29}$$
Plugging any $a > 1$ into this equation will yield a valid value for $n$ and therefore $x$, but they will generally be non-integer.
For $n$ and therefore $x$ to be integer, $4 a^2 + 29$ must be a perfect square, and furthermore an odd perfect square. We can rewrite this as $(2a)^2 + 29 = b^2$, where $a, b \in \mathbb{N}$, which means two perfect squares must have a difference of 29, and this only occurs for $(2a)^2 = 14^2$ and $b^2 = 15^2$. Therefore, solving backwards, $a = 7$, $n = 7$, and $x = 42$.
