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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*}

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    $\begingroup$ What is the question? $\endgroup$ – copper.hat Apr 17 at 23:14
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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.

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  • $\begingroup$ An elegant algebraic explanation! $\endgroup$ – A gal named Desire Apr 17 at 23:48
  • $\begingroup$ Did the solution $(1/2)\sqrt{1+21}$ prompt you to look for the solution to a simpler algebraic expression? $\endgroup$ – A gal named Desire Apr 26 at 15:50
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    $\begingroup$ @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. $\endgroup$ – copper.hat Apr 26 at 17:40
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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

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  • $\begingroup$ My guess is that this was the intended solution. $\endgroup$ – A gal named Desire Apr 17 at 23:35
  • $\begingroup$ What you are really describing is a sequence of real numbers. $\endgroup$ – A gal named Desire Apr 17 at 23:35
  • $\begingroup$ For a proper solution, you would have to say that this sequence is increasing and bounded. $\endgroup$ – A gal named Desire Apr 17 at 23:36
  • $\begingroup$ I guess that for a high school competition, these details would be assumed. $\endgroup$ – A gal named Desire Apr 17 at 23:38
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  • 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

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  • $\begingroup$ How do you know this quartic equation factors into two quadratic polynomials? $\endgroup$ – A gal named Desire Apr 17 at 23:38
  • $\begingroup$ @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 $\endgroup$ – Henry Apr 17 at 23:43
  • $\begingroup$ 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 $5$. $\endgroup$ – A gal named Desire Apr 17 at 23:57
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    $\begingroup$ @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. $\endgroup$ – Henry Apr 18 at 0:03

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