divisible problems 4

$2^{32}-1=(2^{16}-1)(2^{16}+1) \\ Both \quad 2^{16}-1 \ and \ 2^{16}+1 \quad are\ \mathbb{N}atural\ numbers \ greater\ than\ 1\\ Thus\ 2^{32}-1 \quad is\ not\ a\ prime\ number.$

It follows from the fact that if $2^p - 1$ is a prime, then $p$ must be a prime. Since $32$ is clearly not a prime, $2^{32} - 1$ is also not a prime.

NOTE : The converse is NOT true!

No. The function 2^x has the pattern for natural numbers in which the last digit of each output ends in 2, 4, 8, 6 (and continue the pattern.) The number 32 would correspond to the fourth number in this pattern, ie 6. So 2^32 ends in 6. One less this number would end in 5. Therefore 2^32 - 1 is divisible by 5; not prime.

2^32 = 2^4n and last digit of 2^4n is 6 .So last digit of 2^32 -1 = last digit of 2^4n - 1 .i.e. 6-1-5. So it is clear that it is divisible by 5 and hence CAN'T be prime

Note that $2 \equiv -1 \pmod{3}$

Therefore, $2^{32} \equiv (-1)^{32} \pmod{3} \equiv 1 \pmod {3}$

Therefore, $2^{32} -1 \equiv 0 \pmod 3$

$3$ divides $2^{32}-1$ , therefore it is not prime.

2^32 = 4294967296 4294967296-1 = 4294967295 (divisible by 5) 2^32 - 1 is not a prime number

It goes on Mersenne prime format. Where ${ 2 }^{ p }-1$ where $p$ is a prime. Although not all Mersenne numbers are primes, It is safe to assume than anything assuming this format without p being a prime is considered composite number. And since $p=32$ and 32 is not a prime, ${2}^{32} - 1$ is definitely not a prime.

$2^{4}=16\equiv 1\pmod{3}\\ \Rightarrow 2^{32}\equiv 1\pmod{3}\\ \Rightarrow 2^{32}-1\equiv 0\pmod{3}\\ \Rightarrow 2^{32}\ is\ divisible\ by\ 3.\\ \Rightarrow 2^{32}\ is\ not\ a\ prime.$

By Fermats theorm If (x,y) = 1 [co-primes] x^{y-1} -1 is a multiple of y. Hence 2^{33-1} -1 =M[33] Hence its not a prime

This can be written as a difference of squares, a^2-b^2, which in turn can be written as (a+b)(a-b), both of which are integers greater than one in this case; therefore, because it has these factors, it can't be prime.

255 is a factor which is divisible by 3 so not a prime

The final answer will be- 1594884095 and this number is divisible by 5 (because it has the number 5 at the end) and many other numbers. Therefore, it is not a prime number.

2^n - 1 = multiple of 3 when n is even 2^n +1 = multiple of 3 when n is odd

It's just a pattern I saw. I'm too lazy to find a proof for it mathematically.

From the observations if the power of no. 2 is prime no. Then the no. gained which is subtracted by 1 is prime

if we systematically check the terms of the sequence of (2^n)-1, we find that every 2nth term is divisible by 3. Therefore, since (2^32)-1 is the 32nd term and 32= 2*16, (2^32)-1 is divisible by 3 and is not prime.

(2^32 - 1) = (2^16 - 1) * (2^16 + 1) This is a simple difference of two squares.

Its simple really! We know that a^2-b^2 can always be factorized as (a+b)(a-b). This can be extended to any even exponent. Hence in the given problem, a = 2^16 and b=1! Hence the number can be factorized into 2 natural numbers greater than 1 and thus, it isn't a prime number

the power 32 aint a prime so 2^32-1 is not a prime

the given equation can be refactored as (2^16+1)(2^16-1)= (2^2+1)^8 (2^2-1)^8=(5^8)(3^8) which is divisible by 3 and 5 so it is not prime

2^32 = 4294967296 4294967296-1 = 4294967295 (divisible by 5) 2^32 - 1 is not a prime number

Another alternative solution: $2^{1}=2$ (mod 10).. $2^{2}=4$ (mod 10).. $2^{3}=8$ (mod 10).. $2^{4}=6$ (mod 10)..
$2^{5}=2$ (mod 10), so we have a pattern because all the next multiplications will result in the same rest by $10$ . As $32/4$ leaves no remainder $2^{32}$ is congruent to $2^{4}$ in mod 10. So the last digit of $2^{32}-1$ is $5$ , and every number of this kind is divisible by $5$ . Then the aswer is no.

A faster way with much less math is to use the properties of Mersenne numbers.

If n is a composite number, then 2^{n} - 1 is a composite number as well. As 32 has several factors, therefore 2^{32} - 1 will definitely be a composite number.