Let $y^2=x^3+2x+3$ be an elliptic curve over $\Bbb{F}_7$. Find all torsion points of order $2$ and $3$ on $E(\Bbb{F}_7)$.
Now I used the recursive formula to obtain that the third division polynomial of $E$ is $f_3(x)=3x^4+12x^2+36x-4$.
Calculating the torsion points of order $2$ is fairly easy - they correspond to the points for which $y=0$. So taking $y=0$ one can check and see that the points of order two are $\mathcal O,(6,0)$.
But how do we find torsion points of order $3$? Does it have any connection with the third division polynomial? (My professor explained that its roots are the $x$ coordinates of the torsion points of order $3$ but I didn't find this theorem anywhere online).
Every help would be appreciated.
It is fairly easy to compute the third division polynomial explicitly. Assume that $P \in E(\overline{\Bbb F_p})$ is a $3$-torsion point, that is: $3P=O_E$. Then $2P = -P$. If $E : y^2 = x^3+ax+b$, then a well-known doubling formula gives
$$ x(2P) = \dfrac{(3x(P)^2 + a)^2}{4 y(P)^2} - 2 x(P) = x(-P) = x(P) $$
This readily implies that $x(P)$ is a root of $$f_3(x) = (3x^2 + a)^2 - 3x \cdot 4(x^3 + ax + b),$$ which, in our specific case, simplifies into the quartic polynomial $f_3$ that you wrote.
We find that the non-trivial 3-torsion points over $\Bbb F_7$ are $(3, \pm 1)$, and indeed 3 is a root of the polynomial $f_3$.