Exercise 3.2.4 (Numerical examples)

Proof.

(a)

Using the law of quadratic reciprocity, since $5\equiv 1(\mathit{mod}4)$, $$\left(\frac{5}{227}\right)=\left(\frac{227}{5}\right)=\left(\frac{2}{5}\right)=-1.$$

The congruence $x^2\equiv 5(\mathit{mod}227)$ is not solvable.

(b)

Similarly, $$\left(\frac{5}{229}\right)=\left(\frac{229}{5}\right)=\left(\frac{4}{5}\right)=1.$$

The congruence $x^2\equiv 5(\mathit{mod}229)$ is solvable.

(c)

Now $$\left(\frac{-5}{227}\right)=\left(\frac{-1}{227}\right)\left(\frac{5}{227}\right)=(-1)(-1)=1.$$

The congruence $x^2\equiv -5(\mathit{mod}227)$ is solvable.

(d)

Similarly, $$\left(\frac{-5}{229}\right)=\left(\frac{-1}{229}\right)\left(\frac{5}{229}\right)=1\cdot 1=1.$$

The congruence $x^2\equiv -5(\mathit{mod}229)$ is solvable.

(e)

Here, since $1009\equiv 1(\mathit{mod}4)$, $$\left(\frac{7}{1009}\right)=\left(\frac{1009}{7}\right)=\left(\frac{144\cdot 7+1}{7}\right)=\left(\frac{1}{7}\right)=1.$$

The congruence $x^2\equiv 7(\mathit{mod}1009)$ is solvable.

(f)

$$\left(\frac{-7}{1009}\right)=\left(\frac{-1}{1009}\right)\left(\frac{7}{1009}\right)=1\cdot 1=1.$$

The congruence $x^2\equiv -7(\mathit{mod}1009)$ is solvable.