5
$\begingroup$

Is there a non-zero real number $t$ for which there exist infinitely many prime numbers $p$ with $p^{it}$ an algebraic integer?

I would even be surprised to find a real $t \neq 0$ with both $2^{it}$ and $3^{it}$ being algebraic integers.

$\endgroup$
  • $\begingroup$ @Stefan, $2^{2\pi i}=e^{2\pi i\log2}=\cos(2\pi\log2)+i\sin(2\pi\log2)$ and I don't see why that should be algebraic. $\endgroup$ – Gerry Myerson Oct 30 '18 at 4:09
8
$\begingroup$

The six exponentials theorem (look it up) states that if $x_1$ and $x_2$ are two complex numbers which are linearly independent over $\mathbb{Q}$ and if $y_1,y_2$ and $y_3$ are three complex numbers linearly independent over $\mathbb{Q}$ then at least one of the numbers $e^{x_i y_j}$ will be transcendental.

Therefore, for any real number $t$ it is not even possible for three primes $p_1,p_2$ and $p_3$, let alone infinitely many. Take $x_1=1$, $x_2=it$, then take $y_i=\text{log} p_i$ for $i=1,2,3$. Clearly, $y_1,y_2,y_3$ are independent over $\mathbb{Q}$ since $p_i$ are prime numbers and since $t$ is real, $x_1$ and $x_2$ are independt over $\mathbb{Q}$ as well.

The numbers $e^{x_1 y_j}$ are the primes $p_j$ so are algebraic. Therefore one of the numbers $e^{x_2 y_j}=p_j^{it}$ is not algebraic for $j=1,2,3$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I should also comment that the same implication is conjectured for two numbers $y_1$ and $y_2$, this is the four exponentials conjecture. $\endgroup$ – user130124 Oct 30 '18 at 23:33
2
$\begingroup$

https://en.wikipedia.org/wiki/Six_exponentials_theorem

$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$

| cite | improve this answer | |
$\endgroup$
  • 7
    $\begingroup$ Please answer the question in detail. $\endgroup$ – GH from MO Oct 30 '18 at 2:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.