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!


To see what's going on in this problem, it helps to generalize it slightly.  (This should also help put martini's wonderful answer in context.)  Let's look for integers $x$ that make an expression of the form
$$\sqrt{x+\sqrt{x+A}}\in\mathbb{N}$$
for an arbitrary $A\in\mathbb{N}$.
As other answers have shown, setting the expression in question equal to $k$ leads to the quadratic equation
$$x^2-(2k^2+1)x+(k^4-A)=0$$
for which the discriminant is
$$\Delta = (2k^2+1)^2-4(k^4-A)=4k^2+4A+1$$
For $x$ to be an integer, the discrimant must be a square, say $\Delta = n^2$.  This implies
$$(n+2k)(n-2k)=4A+1$$
Now any factorization of the integer $4A+1$, say $4A+1=ab$, gives integer values for $n$ and $k$ (with non-negative $k$ if we take $a\ge b$): Setting  $(n+2k)=a$ and $(n-2k)=b$ implies
$$n={a+b\over2}\quad\text{and}\quad k={a-b\over4}$$
(The denominators here, especially the $4$, may appear problematic, but note that $4A+1$ is congruent to $1$ mod $4$, hence its factors must be odd numbers that are both congruent to either $1$ or $3$ mod $4$.)  Each factorization gives two values for $x$:
$$x={2k^2+1-n\over2}\quad\text{and}\quad x={2k^2+1+n\over2}$$
However, only the first of these is a solution of $\sqrt{x+\sqrt{x+A}}=k$; the other is a solution of $\sqrt{x-\sqrt{x+A}}=k$.
In particular, we always have the factorization $a=4A+1$, $b=1$, and this gives $n=2A+1$, $k=A$, leading to $x=A^2-A$.  This is the solution in martini's wonderful answer.
For $A=7$, the number $4A+1=29$ is prime, so this is the only factorization.  But for $A=11$, for example, we wind up with three values of $x$ that make $\sqrt{x+\sqrt{x+11}}$ an integer:  $x=110$, $x=5$, and $x=-2$, corresponding to the factorizations $45\cdot1$, $15\cdot3$, and $9\cdot5$.
