From Fermat's little theorem we know that every odd prime $p$ divides $2^a-1$ with $a=p-1$.

Is it possible to prove that there are infinitely many primes not dividing $2^a+2^b-1$?

*(With $2^a,2^b$ being incoguent modulo $p$)*

**1.** Obviously, If $2$ is not a quadratic residue modulo $p$ then we have the solution

$a=1, b=\frac{p-1}{2}$

**2.** If $2$ is a quadratic residue and the order of $2 \ modp$ is $r=\frac{p-1}{2}$

then the set $ \{2^1,2^2,...,2^{\frac{p-1}{2}}\}$ is a complete quadratic residue system modp.

So,In this case, $p\mid2^a+2^b-1$ is equivalent to $p\mid x^2+y^2-1$ with $x^2,y^2$ being incogruent modp, which is always true for every $p\geq11$ .

**3.** It is not true that if $p \mid2^a+2^b-1$ and $q\mid2^{a'}+2^{b'}-1$ then $p\cdot q\mid 2^c+2^d-1$ .

There is the counterexample: $5\mid 2^1+2^2-1$ and $17\mid 2^1+2^4-1$ but

$5\cdot 17=85\not \mid2^a+2^b-1$.

We can see a few examples of numbers which have the questioned property :$3,7,31,73,89,...$

(In fact,every Mersenne prime does not divide $2^a+2^b-1$)

Thanks in advance!