Largest root as exponent goes to $+\infty$

If we set $$ p_a(x) = x^{a+2}-x^{a+1}-1 $$ we may easily see that $p_a(x)$ is negative on $[0,1]$, increasing and convex on $[1,+\infty)$, so the largest real root is in a right neighbourhood of $x=1$. We may also notice that:

$$ p_a\left(1+\frac{\log(a+1)}{a+1}\right) = \frac{\log(a+1)}{a+1}\left(1+\frac{\log(a+1)}{a+1}\right)^{a+1}-1>0 $$ by Bernoulli's inequality, hence the largest root of $p_a$ is between $1$ and $1+\frac{\log(a+1)}{a+1}$.

A more effective localization can be achieved by performing one step of Newton's method with starting point $x=1+\frac{\log(a+1)}{a+1}$.

There is a different approach that I have found many times useful: the conversion of a single equation into a system of 2 cartesian equations, here:

$$x^{a+2}-x^{a+1}-1=0 \ \ (E) \ \ \ \ \Longleftrightarrow \ \ \ \ x \ \text{is solution to} \ \begin{cases}y=f_1(x)=x^{a+1}\\y=f_2(x)=\dfrac{1}{x-1}\end{cases}$$

(for a certain $y$). This kind of unfolding (introduction of a supplementary dimension given by ordinate $y$) "geometrizes" in a certain manner the issue.

Let us have a simultaneous look at functions $f_1$ and $f_2$ and at their curves (see below), mainly on $(1,2]$.

• $f_1$ is a ($C^{\infty}$) power function whose U-shaped curve (named hereafter $C_1$) becomes more and more steepy as $a$ increases. More precisely,

$$f_1 \ \text{is strict.} \nearrow, \ \ f_1(1)=1, \ \ \lim\limits_{x \rightarrow\infty} f_1(x)=+\infty \ \ \ \ (1)$$

• $f_2$ is a ($C^{\infty}$) homographic function (independent of parameter $a$) whose curve is a hyperbola ; we only consider its right branch $C_2$ with a vertical asymptote at $x=1$.

$$f_2 \ \text{is strict.} \searrow, \ \ f_1(2)=1, \ \ \lim\limits_{x \rightarrow 1} f_2(x)=+\infty \ \ \ \ (2)$$

Using (1) and (2), it is clear that $C_1 \cap C_2$ has a single point with an abscissa in interval $(1,2]$.

Let us prove in a rigorous manner that, moreover, this point is closer and closer to the upper right part of the asymptote, its abscissa being thus closer and closer to 1.

Let us consider that $a$ takes real positive values (instead of only integer values); with such an extension, every point of $C_2$ is the intersecting point of the hyperbola with one of the curves $C_1$; in fact, if we fix $x_i$ (abscissa of intersection point), the corresponding ordinate is $y_i=\dfrac{1}{x_i-1}$. It suffice then to have $a+1$ such that $y_i=x_i^{a+1}$, i.e., taking logarithms,

$$a=\dfrac{\ln(y_i)}{\ln(x_i)}-1=-\dfrac{\ln((x_i-1)x_i)}{\ln(x_i)}$$

Thus the root can be arbitrarily close to 1, by taking $a$ sufficiently large (if one wants integer solutions, take the "ceiling" of $a$).

Remark: it would be almost immediate to show, by reasoning on $(1,\infty)$ that this root is the largest one of equation $(E)$.

I will show, by elementary means that the root is beteween $1+\dfrac1{\sqrt{a+1}}$ and $1+\dfrac1{a+1}$.

Since the root of $x^{a+2}-x^{a+1}-1 $ is close to $1$, let $x = 1+y$.

Then we want $1 =(1+y)^{a+2}-(1+y)^{a+1} =(1+y)^{a+1}(1+y-1) =y(1+y)^{a+1} $.

Since $(1+y)^{a+1} \gt 1+y(a+1) $, $1 \gt y(1+y(a+1)) \gt y^2(a+1) $ so $y \lt \dfrac1{\sqrt{a+1}} $.

Since $(1+1/(a+1))^{a+1} < e $, if $y < 1/(a+1) $ then $(1+y)^{a+1} < e $ so that $y(1+y)^{a+1} < ey <\dfrac{e}{\sqrt{a+1}} < 1 $ for $a > 8$. Therefore $y > \dfrac1{a+1}$.