To supplement existing data, solutions of
a
p
−
1
≡
1
(
mod
p
2
)
a^{p-1} \equiv 1 \pmod {p^2}
are tabulated for primes
a
,
p
a, p
with
100
>
a
>
1000
100 > a > 1000
and
10
4
>
p
>
10
11
10^4 > p > 10^{11}
. For
a
>
100
a > 100
, five new solutions
p
>
2
32
p > 2^{32}
are presented. One of these,
p
=
188748146801
p = 188748146801
for
a
=
5
a = 5
, also satisfies the “reverse” congruence
p
a
−
1
≡
1
(
mod
a
2
)
p^{a-1} \equiv 1 \pmod {a^2}
. An effective procedure for searching for such “double solutions” is described and applied to the range
a
>
10
6
a > 10^6
,
p
>
max
(
10
11
,
a
2
)
p >\max \, (10^{11}, a^2)
. Previous to this, congruences
a
p
−
1
≡
1
(
mod
p
r
)
a^{p-1} \equiv 1 \pmod {p^r}
are generally considered for any
r
≥
2
r \ge 2
and fixed prime
p
p
to see where the smallest prime solution
a
a
occurs.