Real part of complex root of $X^5-X-k^5+k+1$ is $\frac{1}{23k}$ away from any integer ($k\geq 2$)

Question

Let $k\geq 2$ be an integer. The polynomial $P_k=X^5-X-k^5+k+1$ is easily seen to have exactly one real root and two pair of conjugate non-real roots. Is it true that if $\alpha_k + i \beta_k$ any non-real root of $P_k$, then $\lbrace \lbrace \alpha_k \rbrace \rbrace \geq \frac{1}{23k}$, where $\lbrace \lbrace \rbrace \rbrace$ denotes distance to the nearest integer ? I have checked this for $2 \leq k \leq 10^4$.

My thoughts : $\alpha_k$'s minimal polynomial can be shown to be of degree $10$, so Liouville's approximation theorem would suggest that $\lbrace \lbrace \alpha_k \rbrace \rbrace \geq \frac{C}{k^{10}}$ which is much weaker.

Motivation : I noticed that the real root of $P_k$ is very close to its nearest integer while the real parts of the other roots are comparatively much farther. If quantified successfully, this would yield a proof of the irreducibility of $P_k$ over the rationals.

Answer 1

Not a full solution, but for large $k$ this should explain it.

Let $X = r e^{i \phi}$. Then $$ P_k=r^5 e^{5 i \phi}-r e^{i \phi}- k^5+k+1 $$ Let $a = r^{-4}$.Then
$$ P_k=r^5 (e^{5 i \phi}-a e^{i \phi})- k^5+k+1 $$ For the roots, the imaginary part of this has to vanish, so we have $$ 0 = \sin(5 \phi)-a \sin(\phi) $$ Using the trig. identity $\sin(5 \phi) = \sin(\phi) (16\cos^4(\phi) - 12\cos^2(\phi) + 1)$ this gives $$ 0 = 16\cos^4(\phi) - 12\cos^2(\phi) + 1 -a $$ Again, let $y = \cos^2(\phi)$, reducing this to $$ 0 = 16 y^2 - 12 y + 1 -a $$ This gives $$ y = \cos^2(\phi) = \frac{3\pm \sqrt{4a + 5}}{8} $$

Turning now to the real part of the root, we have

$$ 0 =r^5 (\cos(5 \phi)-a \cos(\phi))- k^5+k+1 $$

With another trig. identity, $$\cos(5 \phi) = \cos(\phi) (16\cos^4(\phi) - 20\cos^2(\phi) + 5) $$ we have, using results from above, $$ 0 =r^5 \cos(\phi) (16\cos^4(\phi) - 20\cos^2(\phi) + 5 -a)- k^5+k+1 \\ = r^5 \cos(\phi) (16y^2 - 20y + 5 -a)- k^5+k+1 \\ =r^5 \cos(\phi) ( - 8y + 4 )- k^5+k+1 \\ =r^5 \cos(\phi) ( 1\pm \sqrt{4a + 5})- k^5+k+1 $$

But also $$ 0 =r^5 \cos(\phi) ( - 8y + 4 )- k^5+k+1 \\ = 4 r^5 \cos(\phi) (1 - 2 \cos^2(\phi))- k^5+k+1 \\ = - 4 r^5 \cos(\phi) \cos(2 \phi)- k^5+k+1 \\ $$ and from above $$ - 4 \cos(2 \phi) = 1\pm \sqrt{5 + 4/r^4} $$

For very large $k$, we have also very large $r$, which gives approximately $$ - 4 \cos(2 \phi) = 1\pm \sqrt{5} \rightarrow \phi = \pm \frac{2 \pi}{5} $$ So $- 4 \cos(\phi) \cos(2 \phi) = 1$ and $$ 0 = r^5 - k^5+k+1 \rightarrow r = k $$ Hence $$ a_k = r \cos(\phi) = \frac{k}{4} (\sqrt5 -1) $$ is the real part of the root for very large $k$. Now, for large $k$, we have that $a_k$ will not be close enough ($< 1/(23 k)$) to any integer, say $n$, since this would approximately require $(4 n +k)^2 = 5 k^2 $.

The same can be done for the other solution pair.

It remains to be estimated how large $k$ has to become such that the derived bounds hold. For smaller $k$, then, validity can be shown manually.