1. ## Solving for x in a quadratic congruence modulo problem

So our class was given this to look over before our last class before the break:
$5x^2 + 5x + 22 \equiv 0mod11$
I got a value that works, $x=11$
but not sure if this is the only answer and I went about it probably not in the correct fashion.

What I did was just notice that $22 \equiv 0mod11$, then I though that if $x=11$, then $55 \equiv 0mod11$, and it turns out that $605 \equiv 0mod11$ as well.
I doubt you can just eliminate terms like this to make it work.

All the examples I've read here don't deal with anything more than say $3x \equiv 1mod4$ or whatever. So I wasn't sure how to approach when a whole quadratic equation was on the LHS.
Also, the quadratic formula doesn't give any real solutions, so my original theory of using that doesn't work.

Any help would be appreciated, Thanks.

2. $5x^2+5x+22\equiv{5x^2+5x}(\bmod.11)$ since $11|22$

So we want $5x^2+5x=5\cdot (x^2+x)\equiv{0}(\bmod.11)$

Since 5 is coprime to 11 it must be that $(x^2+x)=x\cdot (x+1)\equiv{0}(\bmod.11)$

Now, since 11 is prime, it follows that it divides the product if and only if it divides one of the 2 factors.

So the solutions are: $x\equiv{0}(\bmod.11)$ and $x\equiv{-1}(\bmod.11)$

3. Oh, so we're not solving for particular values of x?

Think I understand everything except for what the co-prime means, don't think we've covered it yet.

Thanks, though, now I can go into class knowing a bit more about what to do.

4. Originally Posted by Th3sandm4n
Think I understand everything except for what the co-prime means, don't think we've covered it yet.
It just means if $ab\equiv 0(\bmod n)$ and $(a,n)=1$ then $b\equiv 0(\bmod n)$. If we did not know that $(a,n)=1$ we cannot conclude anything further, as the next example illustrate. Let $a=2,b=3,n=6$ and notice that $ab\equiv 0(\bmod n)$ however, $a\not \equiv 0(\bmod n), ~ b\not \equiv 0(\bmod n)$.