Demonstrate:
If the number $M_p=2^p-1$ is Composite number, where $p$ is prime, then $M_p$ is a Pseudoprime.
This exercise was on a test and I could not do!!
Number Pseudoprime: Fermat's little theorem states that if p is prime and a is coprime to p, then $a^{p−1} − 1$ is divisible by p. For an integer a > 1, if a composite integer x divides $a^{x−1} − 1$, then x is called a Fermat pseudoprime to base a. It follows that if x is a Fermat pseudoprime to base a, then x is coprime to a. Some sources use variations of this definition, for example to only allow odd numbers to be pseudoprimes
More precisely, $M_p$ is a pseudoprime to the base $2$. To show this we show that $$2^{M_p-1}\equiv 1\pmod{M_p}.$$ By Fermat's Theorem we have $2^{p-1}\equiv 1\pmod{p}$. Thus $2^{p-1}=1+kp$ for some integer $k$, and therefore $M_p-1=2kp$. Thus $$2^{M_p-1}=(2^p)^{2k}=(1+M_p)^{2k}\equiv 1\pmod{M_p}.$$