# Solution

(related to Problem: Applications of the Jacobi Symbol)

• $443$ is a prime number, and the solvability of the quadratic congruence can be decided by the calculating the Legendre symbol $\left(\frac{383}{443}\right).$
• In order to do this, we can use the properties of the Jacobi symbol: $$\begin{array}{rcll} \left(\frac{383}{443}\right)&=&-\left(\frac{443}{383}\right)&\text{reciprocity law}\\ &=&-\left(\frac{60}{383}\right)&\text{equal residues}\\ &=&-\left(\frac{2}{383}\right)^2\left(\frac{15}{383}\right)&\text{definition of the Jacobi symbol}\\ &=&-\left(\frac{15}{383}\right)&\text{second supplementary law}\\ &=&\left(\frac{383}{15}\right)&\text{reciprocity law}\\ &=&\left(\frac{8}{15}\right)&\text{equal residues}\\ &=&\left(\frac{2}{15}\right)^3&\text{definition of the Jacobi symbol}\\ &=&1&\text{second supplementary law}\\ \end{array}$$
• It follows that the quadratic congruence $x^2\equiv 383\mod 443$ is solvable.

• $87$ is not a prime number, however, $35$ and $87$ are co-prime.
• We use again the properties of the Jacobi symbol: $$\begin{array}{rcll} \left(\frac{35}{87}\right)&=&-\left(\frac{87}{35}\right)&\text{reciprocity law}\\ &=&-\left(\frac{17}{35}\right)&\text{equal residues}\\ &=&-\left(\frac{35}{17}\right)&\text{reciprocity law}\\ &=&-\left(\frac{1}{17}\right)&\text{equal residues}\\ &=&-1\\ \end{array}$$ 