primitive roots problem. that integer n can never have exactly 26 primitive roots. Show that no integer $n$ can have exactly 26 primitive roots. 
I know that if $n$ has primitive roots then it has exactly $\phi(\phi(n))$ primitive roots.
I think the proof has to use contradiction. 
Suppose $n$ is an integer and has exactly 26 primitive roots then $\phi(\phi(n))=26$. 
How do I carry on and show that $\phi(\phi(n))$ is not 26. 
Please explain me. 

Because 26 is what is called a "nontotient." With only two exceptions, $\phi(n)$ is never odd. There are also some even values that never occur as $\phi(n)$, and 26 is one of them.
Remember that $\phi(p) = p - 1$ if $p$ is prime. But $27 = 3^3$, so 26 is not a totient this way. How about $\phi(p^\alpha) = (p - 1)p^{\alpha - 1}$? Or some combination of primes and powers of primes? Since $26 = 2 \times 13$, we're looking for $n$ a multiple of 3 or 4. But what to make of the 13? If you're still not convinced, you can try testing  $\phi(n)$ for $26 < n < 162$ by brute force, e.g., Select[Range[27, 161], EulerPhi[#] == 26 &].
So, if $\phi(n) = 26$ has no solution, then $\phi(\phi(n)) = 26$ is absolutely hopeless.
