Question: Find the smallest non-negative integer solution (if any exist) to the equation
$$ x^{2017^2} + x^{2017} + 1 \equiv 0\mod\ 2017 $$
Hint: You may use the fact that 2017 is a prime number. Fermat’s Theorem can help with this problem.
I have attempted to apply some addition rules to get
$$ x^{2017^2} \equiv 0\mod 2017 $$
$$ x^{2017}\equiv 0\mod\ 2017 $$
$$ 1 \equiv 0\mod\ 2017 $$
and have gotten stuck. I have also tried to think about how $ x^{2017^2} + x^{2017} + 1 = 2017n $ to satisfy the "$ 0\mod \ 2017$" portion but cannot get much further. I also am not quite sure how to successfully utilize Fermat's Theorem to solve this. I played around a bit but didn't get far.
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