Solution to $\sqrt{\sqrt{x + 5} + 5} = x$

There are natural numbers $$a$$, $$b$$, and $$c$$ such that the solution to the equation $$\begin{equation*} \sqrt{\sqrt{x + 5} + 5} = x \end{equation*}$$ is $$\displaystyle{\frac{a + \sqrt{b}}{c}}$$. Evaluate $$a + b + c$$.

I am not sure where I saw this problem. My guess is that it was from a high school math competition. The solution to the equation is $$\frac{1 + \sqrt{21}}{2}$$. This suggests use of the quadratic formula.

The solution set to the given equation is a subset of the solution set to $$\begin{equation*} x^{2} - 5 = \sqrt{x + 5} , \end{equation*}$$ $$\begin{equation*} x^{4} - 10x^{2} + 25 = x + 5 \end{equation*}$$ $$\begin{equation*} x^{4} - 10x^{2} - x + 20 = 0 . \end{equation*}$$ Using the quartic equation (or Wolfram), the solutions to this equation are computed to be $$\begin{equation*} \frac{1 \pm \sqrt{21}}{2} , \qquad \frac{-1 \pm \sqrt{17}}{2} . \end{equation*}$$

• What is the question? – copper.hat Apr 17 at 23:14

Note that if $$x = \sqrt{x+5}$$ then $$x = \sqrt{\sqrt{x+5}+5}$$.

So, try solving $$x = \sqrt{x+5}$$. This is a quadratic.

• An elegant algebraic explanation! – A gal named Desire Apr 17 at 23:48
• Did the solution $(1/2)\sqrt{1+21}$ prompt you to look for the solution to a simpler algebraic expression? – A gal named Desire Apr 26 at 15:50
• @AgalnamedDesire: No, I let $f(x) = \sqrt{x+5}$ and stated looking for fixed points $x=f(f(x))$ and then thought why not look for a fixed point $x=f(x)$ first. – copper.hat Apr 26 at 17:40

use the following way $$x=\sqrt{5+\sqrt{5 + x} }$$ $$x=\sqrt{5+\sqrt{5 + \sqrt{5+\sqrt{5 + \sqrt{5+\sqrt{5 + ....} }} }} }$$ or $$x=\sqrt{5+x }$$ $$x^2-x-5=0$$ $$x=\frac{1}{2}\pm\frac{\sqrt{21}}{2}$$ now use long division to get the other roots and then check which which one satisfies the original equation

• My guess is that this was the intended solution. – A gal named Desire Apr 17 at 23:35
• What you are really describing is a sequence of real numbers. – A gal named Desire Apr 17 at 23:35
• For a proper solution, you would have to say that this sequence is increasing and bounded. – A gal named Desire Apr 17 at 23:36
• I guess that for a high school competition, these details would be assumed. – A gal named Desire Apr 17 at 23:38
• You can get the four apparent solutions from $$x^{4} - 10x^{2} - x + 20 = (x^2 - x - 5) (x^2 + x - 4)$$ and then solving two quadratic equations

• Squaring can produce spurious results:

• for example if $$x=\frac{-1 - \sqrt{17}}{2}$$ then $$\sqrt{\sqrt{x + 5} + 5} = -x$$ rather than $$+x$$

• but you lost that distinction when you went to $${\sqrt{x + 5} + 5} = x^2$$

• similarly $$x=\frac{-1 + \sqrt{17}}{2}$$ or $$x=\frac{1 - \sqrt{21}}{2}$$ then $$\sqrt{x+5} = -(x^2-5)$$ rather than $$+(x^2-5)$$ but you lost that distinction with the second squaring

When you use transformations which are not $$1-1$$ then you should check any apparent answers in the original question to see whether they are spurious or not

• How do you know this quartic equation factors into two quadratic polynomials? – A gal named Desire Apr 17 at 23:38
• @AgalnamedDesire - (a) empirically it can be factorised (I did) and (b) your stated solutions to the quartic imply that it should be possible to factorise it – Henry Apr 17 at 23:43
• Is this what you're saying? Based on the solution that I gave in the post, you tried to find some "nice" factorization into two quadratic polynomials, and you found it with coefficients of $\pm1$, $4$, and $5$. – A gal named Desire Apr 17 at 23:57
• @AgalnamedDesire - Indeed the solutions in your post implied there would be a nice factorisation of the quartic but it would always be worth looking anyway. – Henry Apr 18 at 0:03