If $a\equiv 1\pmod {p^2},$ then $a\equiv 1\pmod p,$ for $\,a = r^{{\rm ord}(r)}$
$a\equiv\bar a\!\pmod{\!kn}\Rightarrow a\equiv\bar a\!\pmod{\! n};\ $ $(a\bmod kn)\bmod n=a\bmod n.\ $ Congruences persist mod factors of the modulus