solving congruence 12x^2 + 28x + 1 mod 35 How do I solve 12x^2 + 28x + 1 mod 35? I tried breaking it into mod 7 and mod 5 but not sure how to proceed from there.
Below is my sketch work:
$12x^2 + 28x + 1 \equiv 0 \mod 35$ 
$12x^2 + 7x + 1 \equiv 0 \mod 7$ 
$(3x + 1)(4x + 1) \equiv 0 \mod 7$ 
$2x^2 + 3x + 1 \equiv 0 \mod 5$ 
$(2x+1)(x + 1) \equiv 0 \mod 5$ 
 A: Your work so far is great. Now you know that one factor in each equation must be $0$. Since $\mathbb F_5$ and $\mathbb F_7$ are fields, each linear factor has exactly one zero. That gives you two possibilities for each of the remainders, and each of the four combinations gives you one possible remainder $\bmod35$. For instance, using the first factor in each equation, $3x+1\equiv0\bmod7$ yields $x=2\bmod7$, and $2x+1\equiv0\bmod5$ yields $x=2\bmod5$. The unique remainder $\bmod35$ with these remainders is $2$, and indeed $12\cdot2^2+28\cdot2+1=48+56+1=105\equiv0\bmod35$.
A: In $12x^2 + 28x + 1 \equiv 0 \mod 35$, since 28 $\equiv -7 \mod 35$ wouldn't it become $12x^2 - 7x + 1 \equiv 0 \mod 35$? Now, couldn't we just solve this in the reals and then take each of the solutions $\mod 35$? For solutions we would get $1\over 3$ and $1\over 4$, which are 12 and 9, respectively (since $3\times 12\equiv 1$ and $4\times 9\equiv 1 \mod 35$). I checked it and this method does produce answers congruent to 0 mod 35 in the original equation.
