Quadratic Congruence There is a general question whether an integer $a$ has a square root $\pmod{n}$ and if so, how many are there and how one can find them. We know that the solution of the quadratic equation can be found by $$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ If we want to apply this to the case where $a,b,c \in Z_n$, then $2a$ to be unit in $\pmod{n}$, so that we can divide by $2a$. We can write the equation in the form $$(2ax+b)^2=b^2-4ac$$ If we can find all the square roots $s$ of $b^2-4ac$ in $\mathbb{Z}_n$, we can then find all the solutions $x \in \mathbb{Z}_n$ of the quadratic equation of the form $2ax+b=s$, or equivalently $x=(s-b)/2a$. If we look at the square roots at $\mathbb{Z}_{15}$, then 1 and 4 have four square roots each($\pm 1 ,\pm 4$ and $\pm 2,\pm 7$ respectively, while the other units have none. Example: Find all the solutions of $x^2-3x+2 \equiv 0 \pmod{15}$ given $a=1, b=−3$ and $c=2,$ we have: $b^2−4ac=(−3)^2−4(1)(2)=9−8=1$ So, $b^2−4ac=...
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