<s>[INST] Find all solutions of congruence $3x^2−2x+9≡0\pmod {35}$ Find all solutions of congruence $3x^2 - 2x + 9 ≡ 0 \bmod 35$:
Attempt:
\begin{align}
3x^2 - 2x + 9 &\equiv 0 \bmod 35\tag{* 3} \\
9x^2-6x+27 &\equiv 0 \bmod 35 \tag{- 26} \\
(3x-1)^2 &\equiv -26 \bmod 35 \\
\\
-26 + 35 &= 9 = 3^2 \\
\\
\iff (3x-1)^2 &\equiv 3^2 \bmod 35 \\
\iff (3x-1-3)*(3x-1+3) &\equiv 0 \bmod 35 \\
\\
\implies \underbrace{3x - 4 \equiv 0 \bmod 35}_{(a)} 
      &\lor \underbrace{3x + 2 \equiv 0 \bmod 35}_{(b)}\\
\end{align}
Case $(a){:}\; 3x - 4 \equiv 0 \bmod 35$ $\Rightarrow 3x \equiv 4 \bmod 35$ $\Rightarrow 3x \equiv 39 \bmod 35$
$\quad\underset{ ( gcd(3,35) = 1)}{\implies} x \equiv 13 \bmod 35$
Case $(b){:}\; 3x +2 \equiv 0 \bmod 35$ $\Rightarrow 3x \equiv -2 \bmod 35$ $\Rightarrow 3x \equiv 33 \bmod 35$
$\quad\underset{ ( gcd(3,35) = 1)}{\implies} x \equiv 11 \bmod 35$
So $x = 13$ or $x = 11$.
Is it correct that way?
 [/INST] At no point in your proof did you split the congruence into $\bmod5$ and $\bmod7$, so you missed two solutions as I will show below:
$$3x^2-2x+4=0\bmod5\implies x^2+x+3\equiv0\bmod5\implies x\equiv\{1,3\}\bmod5$$
$$3x^2-2x+2\equiv0\bmod7\implies x^2+4x+3\equiv0\bmod7\implies x\equiv\{-1,-3\}\bmod7$$
Combining with the Chinese remainder theorem this gives $x\equiv\{6,11,13,18\}\bmod35$.
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