
> 
> 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.



Algebraic exponential values
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 $j=1,2,3$.


