Number of solutions of $3x^2 - 5x + 3\equiv 0 \pmod{m}$? I'm asked, for each of the following values of $m$, to find the number of solutions (in the set $Z_m$) of the quadratic congruence $3x^2 - 5x + 3\equiv 0 \pmod{m}$. For


*

*$m=53$  

*$m=73$  

*$m=91=7\times 13$ 

*$m=93=3\times 31$ 

*$m=121= 11^2$ 

*$m=527 = 17\times 31$  

*$m=961= 31^2$ 

*$m=2139= 3\times 23 \times 31$.

 A: We can complete the square as usual in the real numbers:
$$3x^2-5x+3=3\left[\left(x-\frac 56\right)^2+\frac {11}{36}\right]$$
This works as written in $\mathbb Z_m$ for any $m$ relatively prime to $36$: i.e. $m$ not divisible by $2$ or $3$. (This is true for all cases in your question except #4 and #8). For those $m$, setting the expression to zero and solving gives
$$\left(x-\frac 56\right)^2 \equiv -\frac{11}{36} \pmod{m}$$
$36$ is always a square number, so the modular equation has two solutions if $-11$ is a square number modulo $m$ and no solutions otherwise. (I suppose I should add that there is one solution if $-11$ is zero, but this is not the case in any of your problems.) This can be tested by using the Legendre symbol and Quadratic Reciprocity.
If $m$ is divisible by $3$ but not by $2$ or $3^2$ (as is the case for problems #4 and #8), dividing the modular equation by $3$ means also dividing the modulus by $3$. The rest of the solution continues as is. So for those cases, the modular equation has two solutions if $-11$ is a square number modulo $\frac m3$ and no solutions otherwise.

Here is how to do #1: Is $-11$ a square modulo $53$?
$$\left( \frac{-11}{53}\right)=\left( \frac{-1}{53}\right)\left( \frac{11}{53}\right)$$
$$=1\left( \frac{53}{11}\right)=\left( \frac{-2}{11}\right)=\left( \frac{9}{11}\right)=1$$
So #1 has two solutions. A quick search using Excel confirms that there are two solutions, $5$ and $32$, but your question did not ask for the actual solutions.
