Euler Totient of Fermat Numbers

Question

I was working on a problem which asks me to prove that the Euler Totient of a Fermat number is always a perfect square.

I found out that every factor of the Fermat number is of the form, $\ k.2^{2^{n}} + 1$

Now, I can use the $n \cdot$ (product of $(1-1/p))$ property to calculate $\phi(n)$, but I have no idea about what k can be.

So, I am going in the wrong direction in order to prove this?

Answer 1

Fermat numbers are of the form $F_n = 2^{2^n}+1$. When they are prime numbers, then $\Phi(F_n) = 2^{2^n}$, which is a perfect square. However...

$F_5 = 2^{2^5}+1 = 4,294,967,297 = 641 \times 6,700,417$

\begin{align} \Phi(F_5) &= \Phi(641) \Phi(6700417) \\ &= 640 \times 67004176 \\ &= 42882672640 \\ &= 2^{11}×5×59×70979 \end{align}

which is not a perfect square.