find all the $n$ such that $ \phi(n) , \phi(n+1) , \phi(n+2)$ are powers of 2 Find all the natural numbers such that, the regular $n , n+1 , n+2 $ gons are constructible. 
Well this problem can be restated in the following way. Since the construction of the regular n-gon is equivalent to the construction of the primitive root $e^{\frac{2\pi i}{n}}$ , and we know that the degree of this extension over $\Bbb Q$ is $\phi(n)$ , and also knowing that a number is constructible iff it's degree over $\Bbb Q$ is $2^k$ we want all the natural numbers  $n$ such that $ \phi(n) , \phi(n+1) , \phi(n+2)$ are powers of 2. I have no idea how can I do this :S, I only know properties of $\phi(n)$ involving products of coprimes. But here ...

We have $\varphi(a)$ is a power of $2$ iff $a$ is a power of $2$ times a product of distinct Fermat primes. Since the only currently known Fermat primes are $3$, $5$, $17$, $257$, and $65537$, there is only a short list of current possibilities to examine.  The largest obvious triple is $(65535,65536,65537)$. 
It is likely that one can show that there are no additional triples, without trying to settle any open problem about the number of Fermat primes. 
We definitely cannot have any more examples of shape $2^{2^n}-1,2^{2^n}, 2^{2^n}+1$ beside the obvious ones. The fact that $F_5$ is not prime settles the case $n=5$. And if $n \gt 5$, then among the factors of $2^{2^n}-1$ we find the Fermat number $F_5$, which is not prime.  There also cannot be examples of shape
 $2^{2^n},2^{2^n}+1, 2^{2^n}+2$ beside $n=0$ and $n=1$. For elementary considerations of size show that $2^{2^n-1}+1$ cannot be a product of distinct Fermat primes if $n \gt 1$. 
