Radical of Seven!

Let $x=\sqrt{7-\sqrt{7+\sqrt{7-\cdots}}}$ and $y=\sqrt{7+\sqrt{7-\sqrt{7+\cdots}}}$ . Then, we can say that $x=\sqrt{7-y}$ and $y=\sqrt{7+x}$ .

Squaring both equations: $x^2=7-y$ and $y^2=7+x$ . Then subtract them: $x^2-y^2=-(x+y) \implies (x+y)(x-y)=-(x+y)$ . Since $x+y\neq 0$ , we can divide both sides by $x+y$ :

$x-y=-1 \implies y=x+1$ .

Finally, substitute that in the first equation:

$x^2=7-x-1 \implies x^2=6-x \implies x^2+x-6=0 \\ (x+3)(x-2)=0$

We take the positive solution, since $x>0$ :

$x=\boxed{2}$ .

Let

$y=\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+....}}}}$

$y=\sqrt{7-\sqrt{7+y}}$

$y^2=7-\sqrt{7+y}$

$y^4-14y^2-y+42=0$

$(y-2)(y+3)(y^2-y-7)=0$

As $\text{the answer is positive and an integer}$

so $\boxed{y=2}$

Let

$y=\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+....}}}}$

$y=\sqrt{7-\sqrt{7+y}}$

$y^2=7-\sqrt{7+y}$

$y^4-14y^2-y+42=0$

$(y-2)(y+3)(y^2-y-7)=0$

As $\text{the answer is positive and an integer}$

so $\boxed{y=2}$

As @Otto Bretscher said above the convergence of the corresponding sequence is very important in this problem and it is the real challenging part. Here we are going to try to justify the convergence of the given expression. Let us define the function $f(x)=\sqrt{7-\sqrt{7+x}}$ defined on the interval $[0, 3]$ . We claim that $f([0, 3])\subset [0,3].\:\:\;\;\:\:\;\:(1)$ Let us prove it. The function $f(x)$ has derivative $f '(x)=-\frac{1}{4\sqrt{x+7}\sqrt{7-\sqrt{x+7}}}$ that is negative at any value of $x\in [0, 3].$ Then $f(x)$ is a decreasing function on $[0, 3].$ Therefore, for any value of $x$ in $[0, 3]$ the value $f(x)$ is in $[f(3), f(0)]$ that is a subset of $[0, 3]$ , and then $f(x)\in [0, 3].$ This proves our claim.

Besides that, we can prove, using calculus as well, that $|f '(x)|$ is decreasing on the interval $[0, 3]$ and, therefore, $\max_{x\in [0,3]}|f '(x)|=|f '(0)|=k=\frac{1}{4\sqrt{7}\sqrt{7-\sqrt{7}}}<1.$ Using the Mean Value Theorem, for any $x_1$ and $x_2$ in $[0, 3]$ the following inequality is true $|f(x_2)-f(x_1)|\leq k |x_2 -x_1|\:\:\:\;\;\:\:\:$ Let us represent the composition $f \circ f \circ f \circ...\circ f,$ where the $f$ is repeated $n$ times, by $f_{n}.$ From (1), we obtain that $f_{n}$ is well defined on the interval $[0, 3]$ and $f_{n}([0, 3])\subset [0,3].$ Additionally, it is very easy to prove that for any natural number $n$ $|f_{n}(x_2)-f_{n}(x_1)|\leq k^n |x_2 -x_1|\:\:\:\;\;\:\:\: (2).$ Now let us defined a sequence by the following recurrence $x_n=f_{n}(0).$ Since $f_{n}([0, 3])\subset [0,3],$ we can obtain that $x_n \in [0,3].$ The expression given in the problem is nothing more than $\lim_{n\to \infty}x_n.$ Then we have to prove that $x_n$ is convergent and its limit is $2$ . Indeed, let us consider two integers $m$ and $n,$ where $m>n$ . Using (2) we get $| x_m -x_n| \leq|f_{m}(0)-f_{m-1}(0)|+|f_{m-1}(0)-f_{m-2}(0)|+...+|f_{n+1}(0)-f_{n}(0)|\leq$ $\leq ( k^{m-1}+k^{m-2}+...+k^{n}) f(0)\leq k^n(1+k +k^2+...+k^{m-1-n})f(0)\leq$ $\leq \frac{k^n}{1-k}f(0).$ Since $k<1$ the sequence $(x)_n$ is a Cauchy sequence and then it converges. For definition and properties of Cauchy sequences you can see for example here.

Let us assume that $\lim_{n\to \infty}x_n=c.$ As $x_n \in [0, 3]$ then $c\in [0, 3].$ Now since $x_{n+1}=f_{n+1}(0)=f(f_{n}(0))=f(x_n),$ using the continuity of $f$ and taking limits as $n$ tends to infinity, we obtain that $c=f(c).$ By inspection we can obtain that a solution of this equation is $c=2.$ As $f(x)$ is decreasing on $[0, 3],$ this is the only solution in this interval. In this way we eliminate the necessity of raising both sides to powers and then the appearance of extraneous solutions difficult to discard. Therefore the given expression of the problem converges and its value is 2. This can answer the question of my student @Jesus Rodriguez

I have also good one, let S=sqr( 7+sqr(...) ) , then it implies sqr(7-S)=sqr( 7-sqr(7+7-S) ) , squaring both sides we get: 7-S=7-sqr(14-S) then S=sqr(14-S) ,square again and we get quadratic equation with condition that S>0 because its sqr(7+sqr...). After that we can write it as S^2=14-S => S^2+S-14=0 =>(S-3)(S+4)=0. Obviously because S>0 we can conclude S=3. Now we go back to problem and we can write it as sqr(7-3)=sqr(4)=2 :)