Hello again!
Conjecture: if 2^(p-1) == 1 (mod p^2),
then phi(p^2)+1 = (p-1)p+1 = q prime.
If so, then 2^(q-1) == 1 (mod p^3).
For p = 1093 and 3511 it works!
Is this a double coincidence?!
If not, narrowing the search
to such cases could yield
a third Wieferich prime.
Best,
Tom Ordo
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