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.
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$\begingroup$ @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 MyersonCommented Oct 30, 2018 at 4:09
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2 Answers
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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 $j=1,2,3$.
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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. $\endgroup$– user130124Commented Oct 30, 2018 at 23:33
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https://en.wikipedia.org/wiki/Six_exponentials_theorem
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7$\begingroup$ Please answer the question in detail. $\endgroup$ Commented Oct 30, 2018 at 2:24