# Algebraic exponential values

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.

• @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. – Gerry Myerson Oct 30 '18 at 4:09

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 $$j=1,2,3$$.

• I should also comment that the same implication is conjectured for two numbers $y_1$ and $y_2$, this is the four exponentials conjecture. – user130124 Oct 30 '18 at 23:33

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

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

• Please answer the question in detail. – GH from MO Oct 30 '18 at 2:24