If
[tex]x^{85}\equiv6\mod29[/tex]
then we have
[tex]\dfrac{x^{85}}6\equiv1\mod29[/tex]
By Fermat's little theorem, if [tex]a[/tex] is not divisible by a prime [tex]p[/tex], then
[tex]a^{p-1}\equiv1\mod p[/tex]
This works for [tex]x=6[/tex]; in this case, we would have
[tex]\dfrac{6^{85}}6=6^{84}=\left(6^3\right)^{28}=\left(6^3\right)^{29-1}\equiv1\mod29[/tex]
