Solving Polynomial Congruences
Example: To solve
\[p(x)=2x^2+4x+1 \equiv 0 \; (mod \; 7)\]
try \[x \equiv 0, \; 1, \; 2, \; 3, \; 4, \; 5, \; 6\]
.\[x \equiv 0 \; (mod \; 7) \rightarrow p(0)= 2(0)^2+4(0)+1=1 \equiv 1 \; (mod \; 7)\]
\[x \equiv 1\; (mod \; 7) \rightarrow p(1)= 2(1)^2+4(1)+1=7 \equiv 0 \; (mod \; 7)\]
\[x \equiv 2 \; (mod \; 7) \rightarrow p(2)= 2(2)^2+4(2)+1=17 \equiv 3 \; (mod \; 7)\]
\[x \equiv 3 \; (mod \; 7) \rightarrow p(3)= 2(3)^2+4(3)+1=31 \equiv 3 \; (mod \; 7)\]
\[x \equiv 4 \; (mod \; 7) \rightarrow p(4)= 2(4)^2+4(4)+1=49 \equiv 0 \; (mod \; 7)\]
\[x \equiv 5 \; (mod \; 7) \rightarrow p(5)= 2(5)^2+4(5)+1=71 \equiv 1 \; (mod \; 7)\]
\[x \equiv 6 \; (mod \; 7) \rightarrow p(6)= 2(6)^2+4(6)+1=97 \equiv 6 \; (mod \; 7)\]
The only solutions are
\[x=1, \; 4\]
. 
