# Solution

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Solve $x^{86} \equiv 6 \pmod{29}$

By Fermat's Little Theorem, we know since $29$ is prime, that if $x \not\equiv 0\pmod{29}$, then $x^{28} \equiv 1\pmod{29}$.

Clearly $x \equiv 0\pmod{29}$ is not a solution as $0^{86} \equiv 0\pmod{29}$

Consequently, we may rewrite the left side of the congruence given in the following way $$x^{86} = (x^{28})^3 \cdot x^2 \equiv 1^3 \cdot x^2 = x^2$$

Thus, we really just need to solve $x^2 \equiv 6\pmod{29}$.

Now we can check the rest by hand $\pmod{29}$:

 $1^2 \equiv 1$ $2^2 \equiv 4$ $3^2 \equiv 9$ $4^2 \equiv 16$ $5^2 \equiv 25$ $6^2 \equiv 7$ $7^2 \equiv 20$ $8^2 \equiv 6$ $9^2 \equiv 23$ $10^2 \equiv 13$ $11^2 \equiv 5$ $12^2 \equiv 28$ $13^2 \equiv 24$ $14^2 \equiv 22$ $15^2 \equiv 22$ $16^2 \equiv 24$ $17^2 \equiv 28$ $18^2 \equiv 5$ $19^2 \equiv 13$ $20^2 \equiv 23$ $21^2 \equiv 6$ $22^2 \equiv 20$ $23^2 \equiv 7$ $24^2 \equiv 25$ $25^2 \equiv 16$ $26^2 \equiv 9$ $27^2 \equiv 4$ $28^2 \equiv 1$

So from the table above, $x$ is a solution to $x^{86} \equiv 6 \pmod{29}$ if and only if either $$x \equiv 8\pmod{29} \quad \textrm{ or } \quad x \equiv 21\pmod{29}$$