Determine $\phi(2^{399}+1)$ This is a question I had on an exam so there is no access to calculators or software. I checked the answer using Maple after and as expected, the answer is very large and some of the prime factors are also very large, with over 10 digits. 

Is there any reasonable way to break this down and get an answer of some sort?
Note that $399 = 3 \times 7 \times 19$.  Letting $x = 2^{21}$, we have
$$
2^{21\times 19} + 1 = x^{19} + 1= (x + 1)(x^{18}-x^{17} + \cdots -x + 1)
$$
Letting $z = 2^7$, we have
$$
x+1 = z^3 + 1 = (z+1)(z^2 - z + 1)
$$
So far, we have
$$
2^{399}+1 = (x^{18}-x^{17} + \cdots -x + 1) \times129 \times 16257
$$
which is a very slight improvement.
After getting feedback on this question in class, it appears that there is no feasible way to do this by hand and it can be easily computed using software such as Maple.