Yes, this is true. This is known as Fermat's Little Theorem, which states that if p is a prime number, then for any integer a, the number a^p-a is an integer multiple of p. Therefore, any number of the form 4^p-1 can be factorized into prime numbers.
Dear Wali Mohd Dar, thank you, but my question is about formulae 4^p-1=a*b*...*z, where a, b, ..., z - different prime numbers.
Dear Peter, all factors, except 3, of 4^p-1 are 2np+1. Many thanks.
It looks quite interesting. No need for p being prime; better 2^n or similar; for example
4^4-1 = 255 = 3 × 5 × 17
also
4^8-1 = 65535 = 3 × 5 × 17 × 257
and also for p=4x11
4^44-1 = 309485009 821345068724781055 = 3 × 5 × 17 × 23 × 89 × 353 × 397 × 683 × 2113 × 2931542417
all the factors are prime... It might be dealt with as a conjecture...
Although perhaps it could not be true... but is easy to check up to big numbers.
Unfortunately your answer on p^2 is not a divisor of 4^p-1 is out of problem. We have to prove that q^2 is not a divisor of 4^p-1 for any q=2np+1 divisor of 4^p-1. So problem has not yet solved.
I am sorry for misunderstanding the problem I try to solve. Of course it is about factorizing numbers (4^p-1)/3, p prime. I send you tech file 38K with factoring the numbers for p
Right. I can repeat the question: It is true that natural numbers (4^p-1)/3, p is prime, factor in different prime numbers? In attached file to last answer you can see that f(p)=(4^p-1)/3. I apologize for mistake.