34

For a specific real number $x$, the difference between any two of $x^{1919}$, $x^{1960}$ , and $x^{2100}$ is always an integer. How would one prove that $x$ is an integer?

user72939
  • 341

1 Answers1

3

First, prove that $x$ is rational.

Say $x^{2100} - x^{1919} = a > 0$ and $x^{1960} - x^{1919} = b > 0$.

Obviously, $x$ is algebraic, of degree at most $1960$. We need to show that the $2099$ other complex solutions to the first equation and the $1959$ other solutions to the second are distinct.

Here, it helps to plot the complex numbers $z$ such that $z^{2100} - z^{1960}$ is a positive real, and similarly with $z^{1960} - z^{1919}$. That graph should look like a bunch of concentric "lines".

We have an asymptotic development for the roots :
The roots of $x^{2100} - x^{1919} = a$ are $\alpha + \frac 1 {2100}\alpha^{-180} + \frac {1739}{8820000}\alpha^{-361} + \ldots$ where $\alpha^{2100} = a$.

Similarly, the roots of $x^{1960} - x^{1919} = b$ are $\beta + \frac 1 {1960}\beta^{-40} + \frac{1879}{7683200}\beta^{-81} + \ldots$ where $\beta^{1960} = b$

Looking at the first term removes every candidate root except $10$ : Since $gcd(2100,1960) = 10$, the only possible common arguments for $\alpha$ and $\beta$ are the $10$ multiples of $\pi/5$ : $\alpha^{10}$ and $\beta^{10}$ have to be positive reals if you want the two roots to have a remote chance at being equal.

We are left with the real axis and $8$ pairs of "lines" that look very close together. Looking at the second term of the expansion, we see that it is real, and very quickly, $\beta^{-40}/1960$ is much larger than $\alpha^{-180}/2100$, so unless we pick real roots, those line stay disjoint from each other.

Hence any simultaneous solution to the two equation has to be one of the two real solutions. And again, the negative solutions are $-x + \frac 2 {2100}x^{-180} + \ldots$ and $-x + \frac 2 {1960} x^{-40} + \ldots$, and they are different for $x$ large enough.

Once you know that $x$ has only one conjugate, and is rational, write $x = c/d$ with coprime $c$ and $d$. Since $c^{2100} = c^{1919}d^{181} + ad^{2100}$, $d^{181}$ divides $c^{2100}$, hence if $p$ is a prime factor of $d$, $p$ also divides $c$, which is impossible. Hence $d=1$ and $x$ is an integer.

mercio
  • 51,469
  • What do you mean by "The roots of $x^{2100} - x^{1919} = a$ are $\alpha + \frac 1 {2100}\alpha^{-180} + \dots$"? Assuming real $\alpha$, that series has a unique sum and defines only one number, right? – ShreevatsaR Apr 20 '13 at 04:05
  • @ShreevatsaR : you get the $2100$ roots of $x^{2100} - x^{1919} = a$ by choosing $\alpha$ among the $2100$ different $2100$th roots of $a$. – mercio Apr 20 '13 at 12:39
  • Oh I see, ok... and how do you get that asymtotic series? – ShreevatsaR Apr 20 '13 at 13:44
  • @ShreevatsaR : Put $t = x/\alpha$ and $u = \alpha^{-181}$. The relation becomes $t^{2100} = 1 + t^{1919}u$. You can then solve this with Puiseux series in $u$. Interestingly you obtain roots in $\Bbb C[[u]]$. The series converge for $u$ small enough (which means $a$ large enough). Actually it should converge for any $|a|$ larger than the $a$ value corresponding to $x=(1919/2100)^{1/181}$ (where the derivative of $x \mapsto x^{2100} - x^{1919}$ vanishes). – mercio Apr 23 '13 at 12:32
  • How did you show that $x$ is rational? I think you showed that $x$ is real...? – user58955 Nov 11 '19 at 10:37