Find the value of $x$, $$\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$$

Help guys please, I have tried and I got, $x=-2, x=1$, and I think it's wrong

  • 4
    $\begingroup$ Successively squaring and moving the fours to RHS leads to a equation like $$x^{16}+\cdots+19596=0$$ with integer coefficients. It has no rational roots. It seems hard, if not impossible to solve by hand. Where did you find it? $\endgroup$
    – ajotatxe
    Commented Jul 5, 2015 at 23:12
  • $\begingroup$ BTW, I have plotted it with Google and it seems to have only one solution around $2.3$ $\endgroup$
    – ajotatxe
    Commented Jul 5, 2015 at 23:15
  • 2
    $\begingroup$ What is the context of the problem? $\endgroup$
    – Winther
    Commented Jul 5, 2015 at 23:36

6 Answers 6


Substitute back the expression for x on the left, and repeat. You arrive, eventually, with the infinite expression $$ x = \sqrt{4+\sqrt{4-\sqrt{4+...}}}$$ By symmetry we are inclined to define $$ y = \sqrt{4-\sqrt{4+\sqrt{4-...}}}$$ Ignoring convergence issues (settled by Herschfeld..I believe Poyla has a good proof, but don't have an exact reference at hand) we easily see that $$x^2 = 4+y$$ and $$y^2 = 4 - x$$ This system is easily solved numerically (I didn't try any other means) and we get $x = 2.30277...$ and it is easy to check that this value works.

  • 3
    $\begingroup$ This system yields the equation $x^4-8x+x+12 = 0$ that has four real roots (I think only one of them works for the original equation). The root you found seems to be $\dfrac{1+\sqrt{13}}{2}$. $\endgroup$
    – Darth Geek
    Commented Jul 5, 2015 at 23:55
  • $\begingroup$ quite right. Ought to have added that I discarded the roots of the quartic which didn't satisfy the original equation. $\endgroup$
    – lulu
    Commented Jul 5, 2015 at 23:57
  • 1
    $\begingroup$ Nice solution! From your system one can derive $(x^2 - x - 3) (x^2 + x - 4) = 0$ for which the system is solved analytically by the quadratic formula giving the analytical solution mentioned by Darth Geek above. $\endgroup$
    – Winther
    Commented Jul 6, 2015 at 0:23
  • $\begingroup$ Excellent solution! Ingenious! $\endgroup$ Commented Jul 6, 2015 at 12:04
  • $\begingroup$ Can't we avoid the convergence issues completely by noticing that the original question is just the 2nd iteration when substituting back $x= \sqrt{4+\sqrt{4-x}}$ to itself? which leads to the same quartic? or is this not allowed? $\endgroup$ Commented Jul 6, 2015 at 17:26

$$\begin{align}\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}&=x&\Longleftrightarrow \\ \left(\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}\right)^2&=x^2&\Longleftrightarrow\\ 4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}&=x^2&\Longleftrightarrow\\ \left(\sqrt{4-\sqrt{4+\sqrt{4-x}}}\right)^2&=\left(x^2-4\right)^2&\Longleftrightarrow\\ 4-\sqrt{4+\sqrt{4-x}}&=\left(x^2-4\right)^2&\Longleftrightarrow\\ -\sqrt{4+\sqrt{4-x}}&=\left(x^2-4\right)^2-4&\Longleftrightarrow\\ \sqrt{4+\sqrt{4-x}}&=-\left(x^2-4\right)^2+4&\Longleftrightarrow\\ \left(\sqrt{4+\sqrt{4-x}}\right)^2&=\left(-\left(x^2-4\right)^2+4\right)^2&\Longleftrightarrow\\ 4+\sqrt{4-x}&=\left(-\left(x^2-4\right)^2+4\right)^2&\Longleftrightarrow\\ \sqrt{4-x}&=\left(x^4-8x^2+12\right)^2-4&\Longleftrightarrow\\ \left(\sqrt{4-x}\right)^2&=\left(\left(x^4-8x^2+12\right)^2-4\right)^2&\Longleftrightarrow\\ 4-x &=\left(x^4-8x^2+10\right)^2\left(x^4-8x^2+14\right)^2 && \end{align} $$

Solving $x$ gives us $\frac{1}{2}+\frac{\sqrt{13}}{2}$

  • $\begingroup$ @mathlove What do you mean? $\endgroup$ Commented Jul 6, 2015 at 13:11
  • 3
    $\begingroup$ What I meant was that you have a mistake at the last. So, I edited it. By the way, you cannot use $\iff$ because they are not equivalent. Also, could you write how you got $x=\frac 12+\frac{\sqrt{13}}{2}$? This is not obvious at all. So, you should add how. $\endgroup$
    – mathlove
    Commented Jul 6, 2015 at 16:58



$$f(\bar{x})=x$$ $$\bar{x}=\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}$$ start putting $x=0$ now you have $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-0}}}}\approx 2.29023...$ now do the same things with the previous result and you get: $$\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-2.29023}}}}\approx 2.3026...$$ If you want a better approximation do it again (you'll get $2.302775...$)

  • 1
    $\begingroup$ This works as a convergent solution, because repeated applications of the square root operation map every value in the domain to $1$ in the limit. I find this solution to be a good way to check for the possibility of a solution existing, not for finding the solution in an analytical form. $\endgroup$ Commented Jul 6, 2015 at 12:06

The polynomial generated by @JanEerland's method will yield a bunch of false roots because every time you square you loose sign information. So you need to test all 16 suggested roots with the original equation.

Here's my Mathematica code:

f[x_] := x^2 - 4;
eqn = f[f[f[f[x]]]] + x // Expand;
aa = Solve[eqn == 0, x];

Output (abbreviated):

(*{{x -> 1/2 (1 - Sqrt[13])}, 
{x -> 1/2 (1 + Sqrt[13])}, 
{x -> 1/2 (-1 - Sqrt[17])}, 
{x -> 1/2 (-1 + Sqrt[17])}...   *)

Now test this solution set against the original equation:

Sqrt[4 + Sqrt[4 - Sqrt[4 + Sqrt[4 - x]]]] - x /. aa // N

Output: (* {3.5876, 0., 4.84176, 0.736553, 4.78663, 4.66183, 4.56239, 3.95255, \ 3.81752, 3.48907, 1.08211, 1.01294, 0.650461, -0.0492326, -0.22168, \ -0.246164} *)

Only the second root in the list returns a zero, so the answer is 1/2 (1 + Sqrt[13])


This should be a comment but then too long ...

Picking up on the solution given by Jan Erlang:

$$\left(x^4-8 x^2+10\right)^2 \left(x^4-8 x^2+14\right)^2+x-4=\left(x^2-x-3\right) \left(x^2+x-4\right) \left(x^{12}-24 x^{10}-x^9+228 x^8+16 x^7-1087 x^6-88 x^5+2720 x^4+191 x^3-3380 x^2-136 x+1633\right)$$

Now $x^2-x-3=0$ yields: $\frac{1}{2} \left(1\pm\sqrt{13}\right)$ and $x^2+x-4=0$ yields $\frac{1}{2} \left(-1\pm\sqrt{17}\right)$ (where only positive roots are acceptable). Further $$x^{12}-24 x^{10}-x^9+228 x^8+16 x^7-1087 x^6-88 x^5+2720 x^4+191 x^3-3380 x^2-136 x+1633=0$$ yields $6$ more positive roots.


We can solve this problem completely, without looking at anything larger than a quartic polynomial.

Let $f(x)=\sqrt{4+\sqrt{4-x}}$. The equation we are trying to solve is $f(f(x))=x$.

Observe that $f(x)$ is strictly decreasing from $f(0)=\sqrt6\gt0$ to $f(4)=2\lt4$, which means the equation $f(x)=x$ has a unique solution, which is relatively easy to find:

$$\begin{align} x=\sqrt{4+\sqrt{4-x}} &\implies x^2=4+\sqrt{4-x}\\ &\implies (x^2-4)^2=4-x\\ &\implies x^4-8x^2+x-12=0\\ &\implies(x^2-x-3)(x^2+x-4)=0 \end{align}$$

There are two roots that satisfy $0\le x\le4$:

$$x={1+\sqrt{13}\over2}\quad\text{and}\quad x={\sqrt{17}-1\over2}$$

Only the first of these satisfies the equation $x=\sqrt{4+\sqrt{4-x}}$; the other one, $x=(\sqrt{17}-1)/2$, satisfies the equation $x=\sqrt{4-\sqrt{4-x}}$.

Since $x=(1+\sqrt{13})/2$ satisfies $f(x)=x$, it also satisfies $f(f(x))=x$, so we have found $a$ solution of our equation. To see that there are no other solutions, we can appeal to calculus.

From $f(x)=\sqrt{4+\sqrt{4-x}}$ we have


from which it will turn out to be convenient to note that, for $x\le3$, $|f'(x)|\le|f'(3))|={1\over4\sqrt5}\lt1$

Meanwhile, since $f(x)$ is decreasing on $0\le x\le4$, the function $F(x)=f(f(x))$ is increasing, hence

$$2\lt\sqrt{4+\sqrt{4-\sqrt6}}=f(f(0))\le f(f(x))\le f(f(4))=\sqrt{4+\sqrt2}\lt3$$

(The crude outer bounds, $2$ and $3$, are just there to simplify some of the arithmetic we're about to do.) Thus any solution to $x=f(f(x))$ satisfies $2\lt x\lt3$, which means $f(x)\lt f(2)=\sqrt{4+\sqrt2}\lt3$.

Now suppose the equation $f(f(x))=x$ had two solutions, say $x=a$ and $x=b$ with $a\lt b$. Then by the Mean Value Theorem applied to $F(x)=f(f(x))$, $F'(c)=1$ for some $c\in(a,b)$. But $2\lt a\lt c\lt b\lt 3$ implies $f(c)\lt3$, and we thus see that


which is a contradiction. We thus conclude that $x={1+\sqrt{13}\over2}$ is the only solution to the equation $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-x}}}}=x$


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