= Solution
By <lifting a primitive root to odd prime powers>, it is enough to test primitivity modulo $25$. The positive integers below eight either fail to be units or have orders
$$
\operatorname{ord}_{25}(1)=1,\quad
\operatorname{ord}_{25}(4)=10,\quad
\operatorname{ord}_{25}(6)=5,\quad
\operatorname{ord}_{25}(7)=4,
$$
whereas
$$
\operatorname{ord}_{25}(2)
=\operatorname{ord}_{25}(3)
=\operatorname{ord}_{25}(8)=20.
$$
Therefore the three smallest required integers are
$$
\boxed{2,\ 3,\ 8}.
$$
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