In my lecture notes it says suppose that $(a,p)=1$ , $p$ is an odd prime then the solutions to
$$ax^2+bx+c\equiv0 (modp)$$
are the same as
$$4a(ax^2+bx+c)\equiv 0(modp)$$
Why is this so ?
I had expected that because $(a,p)=1$ then if $p$ is odd it means that we get $(4a,p)=1$ and that in turn it would mean that $4a(a^2+bx+c)\equiv ax^2+bx+c (modp)$ i.e. that $4a$ would be 1 mod p but checking this for $p=3,a=2$ I got that $4a=2mod3$
On
Assume $4a(ax^2+bx+c)\equiv 0\pmod p.$ This means $p$ divides $4a(ax^2+bx+c)$.
By Euclid's lemma, since $p$ is prime, this means $p$ divides $4$ or $a$ or $ax^2+bx+c.$
No odd prime $p$ divides $4$, and $(a,p)=1$ means $p$ does not divide $a$,
so we're left with $p$ divides $ax^2+bx+c$; i.e., $ax^2+bx+c\equiv0\pmod p$.
Conversely, if $ax^2+bx+c\equiv0\pmod p$,
then it should be obvious that $4a(ax^2+bx+c)\equiv0\pmod p$.
Thus, the solutions to $ax^2+bx+c\equiv0\pmod p$
are the same as those to $4a(ax^2+bx+c)\equiv0\pmod p$.
The statement $(a,p) = 1 $ implies that $ a \not\equiv 0 \mod p$. Since $p$ is odd also $2 \not \equiv 0 \mod p$, so $2 \cdot 2 \cdot a = 4a \not \equiv 0 \mod p$.
Then $4a \cdot f(x) \equiv 0 \mod p \iff f(x) \equiv 0 \mod p$.
It is worth noting that we use the primality of $p$ only at the end. We only use the primality of $p$ to say that a product is zero iff one of its factors is zero. This follows since the ring $\Bbb Z_p$ is an integral domain exactly when $p$ is prime.