I have to find a generator for $Z_{p}^*$. The prime number p is $2425967623052370772757633156976982469681$. My prime factors for (p-1) is according to 1 $f_k=(5,457,571,62429281,174394544633,10673859371941)$. Now I have to test any element $a$ if $a^{(p-1)/f_k}$ $mod$ $p$ not equal 1. My problem is that I'm not sure if that is correct. However, I picked $a=3$ and I got the following results:
$3^{2425967623052370772757633156976982469680/5}$ $mod$ $p$ = $1834816820495944983948594618937337262105$
$3^{2425967623052370772757633156976982469680/457}$ $mod$ $p$ $=$ $1755310908361605643991181654366663815864$
$3^{2425967623052370772757633156976982469680/571}$ $mod$ $p$ $=$ $1249657802614605641962070064914027994794$
$3^{2425967623052370772757633156976982469680/62429281}$ $mod$ $p$ $=$ $1960304120481413108888862269550731072388$
$3^{2425967623052370772757633156976982469680/174394544633}$ $mod$ $p$ $=$ $1104170086449502436618567464298869969539$
$3^{2425967623052370772757633156976982469680/10673859371941}$ $mod$ $p$ = $1641621739539218144241729124053312433783$
According to the definition, $a$ would be generator but i'm sure i'm missing something, because I tested 4 and 5 as $a$ and I got similar results.
Edit: As mentioned below, I forgot to use the prime divisor 2. With that it can be shown that 4 and 5 are no generators but 3 is.
I think you missed the prime divisor $2$ of $p-1$. It would not be surprising for $3$ and $5$ to both be generators, as there are many ($\phi(\phi(p))$) such generators, but $4$ is clearly a quadratic residue and so not a generator.