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According with the graph of the functions $f(x)=x^{x-\sqrt x}$ and $g(x)=\sqrt x+1$ the equation $$x^{x-\sqrt x}=\sqrt x+1;\space x\gt 0$$ has the two approximated solutions $x_1\approx0.216$ and $x_2\approx2.618$.

Well, with a simple trick I have got the exact value (closed form) of $x_2$ but I can't do so for $x_1$. (It clearly could be obtained several ways for $x_1$ a value in approximated form like the given $x_1\approx 0.216$).

My question is the following: Is there any closed (exact) form for $x_1$?

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  • $\begingroup$ You don't think it would be helpful to describe the exact value for $x_2$? $\endgroup$
    – Erick Wong
    Commented Feb 9, 2017 at 19:37
  • $\begingroup$ @Erick Wong: I see: is in the tag with "closed-form". Thank you very much. $\endgroup$
    – Piquito
    Commented Feb 9, 2017 at 19:57
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    $\begingroup$ @Erick Wong: With pleasure: $$x_2=\frac{3+\sqrt 5}{2}$$ Thank you for your edition In spite of the fact that I can not see in what has consisted $\endgroup$
    – Piquito
    Commented Feb 9, 2017 at 20:03
  • $\begingroup$ Well, according to Wolfram Alpha, $x_1 \approx 0.2157321648118760281369885$. Inverse Symbolic Calculator (wayback.cecm.sfu.ca/projects/ISC/ISCmain.html) doesn't find any match for this. It matches the positive root of $341x^2 + 5818x - 1271$ to about 12 decimal places, but not further. So highly unlikely that there is a nice algebraic closed form. $\endgroup$
    – Erick Wong
    Commented Feb 9, 2017 at 21:35
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    $\begingroup$ @Piquito I think in general proving the non-existence of closed forms for a transcendental equation is not easy. For instance, we don't know how to prove that there isn't a non-integer $x$ such that $2^x$ and $3^x$ are both integers. However I can make some partial progress which I will write as an answer. $\endgroup$
    – Erick Wong
    Commented Feb 10, 2017 at 7:08

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Here's a partial answer: if $x_1$ is algebraic, then it must be a quadratic irrational of the specific form $\frac14 + \alpha - \sqrt{\alpha}$ for some non-square rational $\alpha>0$.

Proof: Assume $x_1$ is algebraic. Then so is the RHS $\sqrt{x_1}+1$. By Gelfond-Schneider, the LHS is transcendental unless $x_1 \in \{0,1\}$ (which we know isn't the case), or else $x_1 - \sqrt{x_1}$ is rational. Therefore the latter must be true, which means $x_1$ has degree at most $2$. We now eliminate the possibility that $x_1$ is rational.

Suppose that $x_1$ is rational, then since $x_1 - \sqrt{x_1}$ is rational, so too is $\sqrt{x_1}$. Let $\sqrt{x_1} = \frac{a}{b}$ with coprime $a,b >0$. Since we know $x_1$ approximately we can verify by computation that $a,b>1$, and also that the exponent $x_1 - \sqrt{x_1}$ is negative. We thus have

$$\big(\frac{a}{b}\big)^{-Q} = \frac{a}{b} + 1 \implies \big(\frac{a}{b}\big)^{Q} = \frac{b}{a+b},$$

where $Q$ is a positive rational. This is easily seen to be impossible by the rational root theorem or by unique factorization (for instance, consider a prime $p$ dividing $b$, it must appear only on the denominator on the left but only on the numerator on the right).

It follows that $x_1$ must be a quadratic irrational. It is not hard to check that any number in the interval $[0,\tfrac14]$ for which $x_1 - \sqrt{x_1} \in \mathbb Q$ takes the form $\frac14 + \alpha - \sqrt{\alpha}$.

Perhaps someone can extend the argument much much further and eliminate the quadratic irrational case? One can alternatively use computation to prove that $x_1$ cannot be the root of a quadratic whose coefficients are bounded (in magnitude) by $C$ for various values of $C$. The ISC calculation I did above suggests that can probably take $C = 5800$, and integer relation algorithms can take this bound much higher.

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  • $\begingroup$ By the way, I have had a "mirage" with the equality $$\left(\frac{3+\sqrt5}{2}\right)^{\frac{3+\sqrt5}{2}-\sqrt{\frac{3+\sqrt5}{2}}}=\sqrt{\frac{3+\sqrt5}{2}}+1$$ in apparent contradiction with Gelfond-Schneider but I after realized that the exponent in LHS is precisely $1$. Your uncomplete answer deserves to be the best one (the problem is difficult really). Thank you. Best regards. $\endgroup$
    – Piquito
    Commented Feb 10, 2017 at 11:28

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