I'm not too sure how to answer this question here, so if someone could help me out, that'd be great.

Super $$m$$ is odd and $$a$$ and $$m$$ are coprime. Show that

$$ax^{2} + bx + c \equiv 0$$(mod $$m)$$

has an integer solution $$x \equiv r ($$mod $$m)$$ if and only if $$b^{2} - 4ac$$ is a square modulo $$m$$.

Thanks.

Edit: Here is my proposed solution based on the comments:

$$ax^{2} + bx + c \equiv 0 ($$mod $$m) \implies x^{2} + \frac{b}{a}x + \frac{c}{a} \equiv 0 ($$mod $$m)$$.

$$\implies (x + \frac{b}{2a})^{2} - \frac{b^{2}}{4a} + \frac{c}{a} \equiv 0 ($$mod $$m)$$.

$$\implies (x + \frac{b}{2a})^{2} \equiv \frac{b^{2}-4ac}{4a^{2}} ($$mod $$m)$$

$$4a^{2}$$ is always a perfect square mod $$m$$, so $$b^{2} - 4ac$$ must be a perfect square modulo $$m$$.

One more question is, are we allowing rational solutions for $$x$$? Do we have to worry about $$a/c$$ not being an integer, or $$b^2-4ac$$ being divisible by $$4a^2$$?

• Just mimic the proof of the quadratic formula. Oct 6, 2020 at 13:22
• What happens if you attempt to complete the square? Are there any other usual quadratic equation approaches you are aware of that may help here? Oct 6, 2020 at 13:22
• Start by saying $ax^2+bx+c\equiv0$ and, as suggested above, mimic the proof of the quadratic formula, but for integers mod $m$ rather than for real numbers Oct 6, 2020 at 13:32
• @Cjw123 you need to be more careful in what you say. In particular $1/a$ is not a thing in this context. You must note that since $gcd(a, m) = 1$ $a$ has a multiplicative inverse mod $m$. In this way it is better to write "$a^{-1}$" to mean the "inverse of $a$ in the ring $\mathbb{Z}/m\mathbb{Z}$." This way we are not "allowing rational solutions" - we are just never leaving $\mathbb{Z}/m\mathbb{Z}$. You should also notice that you have only shown one direction (although the other should not be hard from where you are) Oct 6, 2020 at 17:52
• Similarly when you "divide by 2" this is admissable since $m$ is odd, so $2$ has a multiplicative inverse mod $m$. These are all things you should be noting in your proof Oct 6, 2020 at 17:53

It's possible to use rational numbers in modular arithmetic, but you have to be careful about what you mean. Instead of what you did, multiply both sides by $$4a$$ and put the $$c$$ on the other side. Then when you complete the square, you won't have fractions.
• Perhaps I'm making a really dumb algebra error, but I'm getting the following: $ax^2 + bx + c \equiv 0 (modm) \implies 4a^2x^2 + 4abx \equiv -4ac (modm).$ When I'm completing the square on $4a^2x^2+4abx$, I'm trying to form it into 4(ax+b)^2 but i'm getting an extra factor of $abx$ on there. I'm doing something wrong....
• @Cjw123 You should form it into $(2ax +b)^2$ instead. Oct 6, 2020 at 14:04