I have read a wikipedia article on Pisot number or PV number: <https://en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number>


I define a Pisot number $\alpha$ is "small" iff $\alpha<2$.(It is not a qualified definition; is there another reasonable bound for the problem below?)


Is there a small Pisot number $\alpha$ such that every (Galois) conjugate of it is real, except for the golden ratio $\frac{1+\sqrt{5}}{2}$? The conjugates of any quadratic Pisot number are real, but it is larger than 2 other than $\frac{1+\sqrt{5}}{2}$.


If then, how many? The nonexistence may be an exercise on Pisot numbers, but I can't even guess how I can approach it.


(The same post in Math Stackexchange was deleted.)



small Pisot numbers with real conjugates
The following theorem of Kronecker seems to (almost) answer your question.


 *Let $\beta$ be an algebraic integer such that all its algebraic conjugates are real (i.e. $\beta$ is totally real). Then either $\beta$ is of the form $\beta=2\cos(\pi q)$ with rational $q$, or some algebraic conjugate of $\beta$ has absolute value strictly greater than 2.*


Due to this theorem, we only need to find for which $n>3$ the number $\beta=2\cos (\pi/n)$ is a Pisot number; in other words, there should be no integers coprime with $2n$ in the intervals $[2,n/3]$ and $[2n/3,n]$. (The golden ratio correspods to $n=5$). 


Thus, if $p$ is the smallest prime not dividing $n$ then $p>n/3$. This leaves us with only small cases which one may deal with by hands.


