How many prime numbers p are there such that 29^p+1 is a multiple of p?
4 answers
querry
how many prime numbers are such that 29^p+1 is a multiple of p?
We have (A) : 29^(p)=-1(mod p) , so 29^(2p) = 1(mod p)
Let k=ord p (29)
k | 2p implies k={1, 2} since k is different from p from (A).
If k=1 we get : 28 = 0(mod p) which means p € {2, 7}.
If k=2 we get : 840=0(mod p) which means p € {2, 3, 5, 7}.
Finally, there are 4 prime numbers.
Let k=ord p (29)
k | 2p implies k={1, 2} since k is different from p from (A).
If k=1 we get : 28 = 0(mod p) which means p € {2, 7}.
If k=2 we get : 840=0(mod p) which means p € {2, 3, 5, 7}.
Finally, there are 4 prime numbers.
By fermat's little theorem, 29^p = 29 mod p so 29^p+1 = 30 mod p,
and we're given that 29^p+1=0modp, therefore 30=0modp, and p | 30
This yields 3 answers, namely p=2,3,5.
and we're given that 29^p+1=0modp, therefore 30=0modp, and p | 30
This yields 3 answers, namely p=2,3,5.