TITLE: Solving Exponential Diophantine Equations
QUESTION [1 upvotes]: Find all positive integers $a, b, c > 1$, all relatively prime with respect to each other such that $b\mid 2^{a} +1, c\mid 2^{b} +1, a\mid 2^{c} +1$. $$$$ It is easy to see that all of $a, b, c$ are odd. Suppose $3\mid a$ then we have $3\mid 2^c+1$ and as $(a, c)=1$ so we have $c=6k+1$ or $c=6k+5$ for some $k$. But $9\mid a$ implies $9\mid 2^c+1$ which is not possible. This is true for all the three $a, b, c$. Hence either all of them are equal to $3$ or have some prime factor other than $3$. Suppose the later one is true.  Let $p$ be a smallest prime factor of $b$ other than $3$ then we have $$2^{2a} \equiv 1 \mod p$$ and also we have $$2^{p-1} \equiv 1 \mod p$$
So we have $$2^d \equiv 1 \mod p$$ where $d=\gcd (2a, p-1)$. Now $d \neq 2$ as $p \neq 3$. Also $d \neq 3$ because that would have implied $p=7$ and $3|a$ and looking modulo $7$ we arrive at a contradiction. So let $q$ be a prime factor other than $3$ dividing both $a$ and $p-1$ then we have $q<p$ and also as $a$ divides $2^c+1$ we have $$2^{2c} \equiv 1 \mod  q$$ and also $$2^{q-1} \equiv 1 \mod q$$. Again by the same reasoning as above we can show that there is a prime factor $r$ other than $3$ dividing both $c$ and $q-1$ and hence we have $r<q$ and as $c|2^b+1$ we have $$2^{2b} \equiv 1 \mod r$$ and also we have $$2^{r-1} \equiv 1 \mod r$$. Again by the same reasoning we can show that there exists a prime $s$ other than $3$ dividing both $b$ and $r-1$ and hence $s<r<q<p$ and $s$ divides $b$ bot as $p$ was the smallest prime factor other than $3$. So a contradiction and hence there do not exist such positive integers.
$$$$Is My Proof Correct?????

REPLY [1 votes]: The reasoning is correct, but not all steps are obvious (to me).
Some remarks:

$3 \mid 2^c+1$ is always true as $c$ is uneven. $2^{2k+1}+1=2.(4)^k+1 \equiv 2.1+1 \equiv 0 \pmod{3}$.
$2^{6k+r}+1=2^r.(64)^k+1 \equiv 2^r.1+1 \pmod{9}$. Hence $9 \mid 2^{6k+r}+1$ implies $r=3$, and thus $3 \mid c$. Hence, assuming $a=3^n$ implies $a=3$ as $(a,c)=1$, and as $b \mid 2^a+1=9$ we conclude that $b=1$. This implies $c=1$. But $(3,1,1)$ is not a feasible solution.
$2^a \equiv -1 \pmod{p}$ implies $2^{2a} \equiv 1 \pmod{p}$
$d=2\gcd(a,(p-1)/2)$. Now $d=6$ implies $2^6 \equiv 1 \pmod{p}$, hence $p=7$. Furthermore, $3 \mid a$ so $a$ is of the form $a=3(2k+1)=6k+3$. Hence $2^a+1=2^{6k+3}+1=$ $8.(2^6)^k+1 \equiv 1.1+1=2 \pmod{7}$. This contradicts with $7 \mid 2^a+1$. We conclude that $d \ge 10$ and $p \ge 11$.

I found an excellent paper with several lemma's regarding $2^n+1$. When looking for all (not coprime) solutions, it is remarkable that only a few prime numbers can appear in $a$, $b$ or $c$; see A057719.
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