<s>[INST] How can I solve quadratic equations using modular arithmetic How can I solve quadratic equations using modular arithmetic? E.g.
$$2x^2 + 8x + 2 = 0 \pmod{23}$$
N.b. I have changed the figures from those in my homework question because I don't want a solution I want to understand the process. Consequently the example I gave might not have solutions. For the example I am working from divide the LHS by 2.
 [/INST] We have $2x^2+8x+2\equiv 0\pmod{23}$ if and only if $x^2+4x+1\equiv 0\pmod{23}$. 
Now complete the square. We have $x^2+4x+1=(x+2)^2-3$. So we want to solve the congruence $(x+2)^2\equiv 3\pmod{23}$.
Let $y=x+2$. We want to solve the congruence $y^2\equiv 3\pmod{23}$.
There is general theory that, for large $p$, helps us determine whether a congruence $y^2\equiv a \pmod{p}$ has a solution, and even to compute a solution. But at this stage you are probably expected to solve such things by inspection. 
Note that $y\equiv 7\pmod{23}$ works, and therefore $y\equiv -7\equiv 16\pmod{23}$ also works. We have found two solutions, and by general theory if $p\gt 2$ there are either $2$ solutions or none, so we have found all the solutions. 
From $y\equiv 7\pmod{3}$ we conclude that $x+2\equiv 7\pmod{23}$, and therefore $x\equiv 5\pmod{23}$.
From $y\equiv 16\pmod{23}$ we conclude that $x\equiv 14\pmod{23}$.
Remarks: If our congruence had been (for example) $x^2+7x-8\equiv 0\pmod{23}$, there would be a bit of unpleasantness in completing the square, since $7$ is odd. But we could replace $7$ by, say, $30$, and complete the square to get $(x+15)^2-225-8$. So our congruence would become $(x+15)^2\equiv 233\pmod {23}$, or equivalently $(x+15)^2\equiv 3\pmod {23}$.
In general, if we have $ax^2+bx+c\equiv 0\pmod{p}$, it is useful to multiply through by the inverse of $a$ modulo $p$ to make the lead coefficient equal to $1$. There are a number of other helpful "tricks."
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