I've just been given the following question in my crypto class, and I think I'm fairly sorted for it, but I was just wondering whether there might be any extra solutions to the ones I've worked out.
Compute all solutions of $x^2 + 4x - 21 \equiv 0\,\bmod\,33$
First, I factorised this equation to give $(x + 7)(x - 3)$, which gives me the solutions $-7$ and $3$. However, under the conditions of modular arithmetic, I know that adding or subtracting $33$ as many times as we like will also provide an answer of zero.
IE: Let's try x = 26. $$(26 + 7)(26 - 3) = 33 \times 23 \equiv 0\,\bmod\, 33$$
Thus, it becomes fairly obvious to see that solutions to this equation will take the form $$ and $$, where the square brackets denote congruence classes.
I was given the hint in class that we should try to make the brackets equal to the factors of 33 - IE; try and get to $11 \times 3$, for example. But I really can't see how this would work.
Any further input would be great, thank you!!