Poly-Q2.md

describe all $n$ such that $(n-2)(n-3) \equiv 0 \pmod{97}$

https://crypto.stanford.edu/pbc/notes/numbertheory/poly.html

Find zero points then because 97 is prime we know the degree 2 polynomial can have at most 2 roots and so we are done.

So solutions are $n \equiv 2 \pmod{97}$ and $n \equiv 3 \pmod{97}$