infinite primes

[asif e elahi]

Prove there are infinite primes $p$ and $q$,$p\neq q$ such that $p\mid 2^{q-1}-1$ and $q\mid 2^{p-1}-1$.

[Nirjhor]

By FLT, any odd prime $p\mid 2^{p-1}-1$. So just consider $p=q$ and infinitude of primes completes the rest.

[asif e elahi]

The problem was wrong.Now it is edited.Here $p$ and $q$ are distinct.

[photon]

What's wrong ?
By Fermat's little theorem and as given , $pq|2^{p-1}-1$ and $pq|2^{q-1}-1$ .
Let $2^{p-1}=pqc+1$ , $2^{p-1}=pqd+1$ . ( $c,d$ are 2 positive odd integers .) WLOG , $p<q$ .
$2^{p-1}+2^{q-1}=2^{p-1}(2^{q-p}+1)$
$\Rightarrow pq(c+d)+2=2^{p-1}(2^{q-p}+1)$ ,
$\Rightarrow pqm+1=2^{p-2}(2^{q-p}+1)$ [$c+d=2m$] ,
$2^{p-2} \parallel pqm+1$ again , $2^{p-1} \parallel pqd+1$ , $\therefore 2^{p-2} \parallel pq(d-m)$
$2^{p-1} \parallel 2pqd-2pqm $ .....(1) , $2^{p-1} \parallel 2pqm+2 $ ......(2)
Therefore , $2^{p-1} \parallel (2pqm+2) +(2pqd-2pqm) =2(pqd+1)$ ,
but $2^{p-1} \parallel pqd+1$ . contradiction . So there are no prime following such property .

[asif e elahi]

$2^{p-2} \parallel pqm+1$ again , $2^{p-1} \parallel pqd+1$ , $\therefore 2^{p-2} \parallel pq(d-m)$
.

How you got $2^{p-1}$ fully divides $pqd+1=2^{q-1}$ ?