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!
If you just want to find some $x$, not all, you can try to find $x$ such that $\sqrt{x+7} = 7$, as then $$ \sqrt{x + \sqrt{x+7}} = \sqrt{x+7} = 7 \in \mathbb N.$$ Can you see such an $x$?
$$x + 7 = ((n^2)-x)^2 = n^4 - 2n^2x + x^2$$ $$x^2-(2n^2+1)x+n^4-7=0$$ $$x_{1,2}=\frac{2n^2+1 \pm \sqrt{(2n^2+1)^2-4(n^4+7)}}{2}\\ =\frac{2n^2+1 \pm \sqrt{4n^4+4n^2+1-4n^4+28}}{2}\\ =\frac{2n^2+1 \pm \sqrt{4n^2+29}}{2}\\$$
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 .
Trial and Error
I assume you want $x$ to be an integer. You might want $x+7$ to be a square so that $\sqrt{x+7}\in\mathbb{N}$... $x+7\in\{1,4,9,16,25,36,49,64\dots\}$ and hence $$\begin{align}x&\in\{-6,-3,2,18,29,42,57,\dots\} \end{align}$$
But you want $x+\sqrt{x+7}$ to be a square so look at the corresponding values of it:
$$x+\sqrt{x+7}\in\{-5,-1,5,22,34,49,65,\dots\}.$$
One of these is squares, corresponding to $x=42$.
$x + 7 = ((n^2)-x)^2 = n^4 - 2n^2x + x^2$ which is quadratic equation considering x. Can you go further?
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 $9\cdot5$.
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 $x = 42$.
just to it by guesswork , x=42
You're mentioning that $x$ should have two digits. Write a small program that runs through all integers from 10 to 99, computes your expression and finds that 42 is the only solution.
Example in Python:
>>> import math
>>> for x in range(10,100):
... n = math.sqrt(x + math.sqrt(x + 7))
... if float(int(n)) == n:
... print("The solution is {0}".format(x))
...
The solution is 42
