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I ran into a problem, and I'm not sure how to continue.

Problem: Let $f$ be a function such that $\sqrt {x - \sqrt { x + f(x) } } = f(x)$, for $x > 1$. In that domain, $f(x)$ has the form $\dfrac{a+\sqrt{cx+d}}{b}$, where $a$, $b$, $c$, $d$ are integers and $a$, $b$ are relatively prime. Find $a+b+c+d$.

So, I tried to cancel out the radicals, and got $(f(x))^4-2x(f(x))^2-f(x)+x^2-x=0$. Setting $y=f(x)$, I tried to apply the quadratic formula to find $x$ in terms of $y$. I got $$ x=\frac{x+2xy^2 \pm \sqrt{(x+2xy^2)^2-4(y^4-y)}}{2}. $$ From here, I tried simplifying the radical, but got $$ x=\frac{x+2xy^2 \pm \sqrt{(x^2)(2y^2+1)^2-y(4y^3+1)}}{2}. $$ I don't know if I factored it wrong, or if I'm missing something painfully obvious. Can I have a hint as to how to continue? It would be greatly appreciated.

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  • $\begingroup$ I think when you square your equality you should obtain: $$ x - \sqrt{x+f} = f^2\\ x^2 - 2x\sqrt{x+f} + x + f = f^4\\ f^4- x^2 + 2x\sqrt{x+f} - x - f = 0 $$ $\endgroup$
    – dustin
    Commented Oct 26, 2014 at 17:38
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    $\begingroup$ Could you explain why? I squared both sides, resulting in $x-\sqrt{x+f(x)}=(f(x))^2$, then $\sqrt{x+f(x)}=x-(f(x))^2$. Squaring again results in $x+f(x) = x^2-2x(f(x))^2+(f(x))^4$. Subtracting $x+f(x)$ from both sides yields $x^2-2x(f(x))^2+(f(x))^4-x-f(x)=0$. $\endgroup$ Commented Oct 26, 2014 at 17:40
  • $\begingroup$ Your algebra eliminating radicals seems to be correct. Writing $y=f(x)$, it's $y^4 - 2xy^2 + x^2 - x - y = 0$, which is a little easier on the eyes. $\endgroup$ Commented Oct 26, 2014 at 17:41
  • $\begingroup$ I made an edit to the question for the substitution. I'm still wondering whether it is possible to simplify the inside of the square root? $\endgroup$ Commented Oct 26, 2014 at 17:46
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    $\begingroup$ Your use of the quadratic formula seems off - you shouldn't be having $x$ terms in the RHS. In any case I am not sure why you would want to have $x$ in terms of $y$. Having $y$ in terms of $x$ is more useful, as you can say $$y = \dfrac{-1\pm \sqrt{4x-3}}2$$ for e.g. and compare to get $a, b, c, d$. $\endgroup$
    – Macavity
    Commented Oct 26, 2014 at 17:53

2 Answers 2

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We have $f(x)=O(\sqrt x)$. Note that $\lim_{x\to\infty}\frac{f(x)}{\sqrt x}=\frac{\sqrt c}{b}$, whereas $\lim_{x\to\infty}\frac{\sqrt{x-\sqrt{x+f(x)}}}{\sqrt x}=1$. So we conclude $c=b^2$. Then $$ \sqrt{x+f(x)}=x-f(x)^2=x-\frac{a^2+2a\sqrt{b^2x+d}+b^2x+d}{b^2}=-\frac{(a^2+d)+2a|b|\sqrt{x+d/b^2}}{b^2}$$ Repeat the trick from above, i.e. divide by $\sqrt x$ and take the limit as $x\to\infty$, to conclude that $-\frac{2a|b|}{b^2}=1$. From this with $\gcd(a,b)=1$, conclude $a=-1$, $b=2$, hence $c=4$. Can you do the last step to find $d$?

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  • $\begingroup$ Can you explain your steps involving limits? I'm not sure I quite follow your thinking... $\endgroup$ Commented Oct 26, 2014 at 17:52
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Putting $y=f(x)$, your equation becomes:

$y^4 - 2xy^2 + x^2 - x - y = 0$

This is hard to solve for $y$, but we have the suggestion that $y$ has the form $y=\frac{a+\sqrt{cx+d}}{b}$. This can be substituted into our polynomial and then simplified:

$\left(\frac{a+\sqrt{cx+d}}{b}\right)^4 - 2x\left(\frac{a+\sqrt{cx+d}}{b}\right)^2 + x^2 - x - \frac{a+\sqrt{cx+d}}{b} = 0$

$\implies\left(a+\sqrt{cx+d}\right)^4 - 2b^2x\left(a+\sqrt{cx+d}\right)^2 +b^4x^2 - b^4x - b^3\left(a+\sqrt{cx+d}\right) = 0$

$\implies a^4+4a^3\sqrt{cx+d}+6a^2(cx+d)+4a(cx+d)^{3/2}+(cx+d)^2-2a^2b^2x-4ab^2x\sqrt{cx+d}-2b^2x(cx+d)+b^4x^2-b^4x-ab^3-b^3\sqrt{cx+d}=0$

$\implies a^4+6a^2(cx+d)+(cx+d)^2-2a^2b^2x-2b^2x(cx+d)+b^4x^2-b^4x-ab^3=b^3\sqrt{cx+d}+4ab^2x\sqrt{cx+d}-4a^3\sqrt{cx+d}-4a(cx+d)^{3/2}$

$\implies a^4+6a^2(cx+d)+(cx+d)^2-2a^2b^2x-2b^2x(cx+d)+b^4x^2-b^4x-ab^3 = (b^3+4ab^2x-4a^3-4a(cx+d))\sqrt{cx+d}$

$\implies (a^4+6a^2(cx+d)+(cx+d)^2-2a^2b^2x-2b^2x(cx+d)+b^4x^2-b^4x-ab^3)^2 = (b^3+4ab^2x-4a^3-4a(cx+d))^2(cx+d)$

If you multiply this all out, and move everything to the left, you'll have a polynomial of the form $g(x)=M_4x^4 + M_3x^3 + M_2x^2 + M_1x + M_0 = 0$, where each of the $M_i$'s is a polynomial in $a$, $b$, $c$ and $d$. Since $g(x)=0$ for all $x>1$, we can say that $M_4=M_3=M_2=M_1=M_0=0$. If there is a solution to this system of five polynomials in four variables, then that's your answer.

In general, systems of polynomials can be difficult, but certain software packages allow for Groebner basis calculations, which can work, depending on the computer's capabilities and the user's patience.

One hopes there's an easier way to solve this, but the above is a sort of brute-force approach.

EDIT The solution posted here by Hagen von Eitzen using limits is considerably nicer.

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