Quote:
Originally Posted by murzyn0
2^p  1, where p is prime number is always prime number, for example:
2^7  1 is 127,
2^127  1 is 170141183460469231731687303715884105727,
2^170141183460469231731687303715884105727  1 is big number, but its prime number, so it's eveidnce that there is infinity mersenne prime numbers

Eveidnce [sic] != proof.
Quote:
Originally Posted by murzyn0
but, 13 in 2^131 is not a mersenne prime numbers.
2^p  1, where p is a mersenne prime, yields a different mersenne prime.

2^51 (=31) is prime. 2^311 is prime. But 2^(2^311)1 is composite, factors are known.
How back to you go? Because 5 is not a Mersenne prime. And your example above, 2 is not a Mersenne prime either, so the sequence 2, 3, 7, 127, ... doesn't start with a Mersenne prime.
And if you conveniently ignore the first term then 3, 7, 127, ... does match your claim, but then 31, 2147483647, ... fails your claim. You can't have it both ways.