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!


$$x + 7 = ((n^2)-x)^2 = n^4 - 2n^2x + x^2$$
$$x^2-(2n^2+1)x+n^4-7=0$$
$$x_{1,2}=\frac{2n^2+1 \pm \sqrt{(2n^2+1)^2-4(n^4+7)}}{2}\\
=\frac{2n^2+1 \pm \sqrt{4n^4+4n^2+1-4n^4+28}}{2}\\
=\frac{2n^2+1 \pm \sqrt{4n^2+29}}{2}\\$$
This is an integer if and only if $ \sqrt{4n^2+29}$ is an integer (the converse follows immediately from the observation that if $\sqrt{4n^2+29}$ is an integer, it must be odd.
Claim 1: If $\sqrt{4n^2+29}$ is an integer, then $n \leq 7$.
Proof:
$$\sqrt{4n^2+29}  > \sqrt{4n^2}=2n \,.$$
Thus
$$\sqrt{4n^2+29}  \geq 2n+1 \,.$$
Hence
$$4n^2+29  \geq 4n^2+4n+1 \,.$$
This implies that $n \leq 7$, with equality if and only if $n=7$.
Claim 2: If $\sqrt{4n^2+29}$ is an integer, and $n \neq 7$ then $n \leq 1$.
Proof: Exactly like in Claim $1$ 
$$\sqrt{4n^2+29}  \geq 2n+1 \,.$$
But we proved that we only have equality for $n=7$. Thus
$$\sqrt{4n^2+29}  > 2n+1 \,.$$
As $\sqrt{4n^2+29}$ is odd, we get
$$\sqrt{4n^2+29}  \geq 2n+3 \,.$$
$$4n^2+29 \geq 4n^2+12n+9 \,.$$
$$29 \geq 12n+9 \,.$$
Thus $n \leq 1$.
Thus we showed that the only $n$ which can work are $n = 0, n = 1$ and $n = 7$.
If 
$$n=0 \Rightarrow x_{1,2}
=\frac{1 \pm \sqrt{29}}{2} \notin \mathbb Z$$
$$n=1 \Rightarrow x_{1,2}
=\frac{2+1 \pm \sqrt{4+29}}{2}\notin \mathbb Z$$
$$n=7 \Rightarrow x_{1,2}
=\frac{99 \pm 15}{2}$$
P.S. Don't forget that we squared couple times, which means that the solutions we got are solutions to $$x + 7 = ((n^2)-x)^2 = n^4 - 2n^2x + x^2$$
but not necessarily to the original question.
One needs to check them in the original equation, and only one works.
The reason why the other solution doesn't work is because it appears as an extra solution when we square:
$$\sqrt{x + 7 }= (n^2)-x$$
If we observe in the original equation that $n^2 \geq x$, this eliminates the wrong extra solution .
