Exercise 2.9.1 (Second degree equation modulo $p$)

Proof.

(a)

$$\begin{array}{llll}\hfill 4x^2+2x+1& \equiv -x^2+2x+1& \hfill & \\ \hfill & \equiv -(x^2-2x-1)& \hfill & \\ \hfill & \equiv -\left[{(x-1)}^2-2\right](\mathit{mod}5).& \hfill & \\ \hfill & & \hfill \end{array}$$

Therefore, for all $x\in \mathbb{Z}$,

$$4x^2+2x+1\equiv 0(\mathit{mod}5)⟺{(x-1)}^2\equiv 2(mod5).$$

Since $2$ is not a square modulo 5, this equation has no solution.

(b)

$$\begin{array}{llll}\hfill 3x^2-x+5& \equiv 3(x^2+2x+4)& \hfill & \\ \hfill & \equiv 3\left[{(x+1)}^2+3\right](\mathit{mod}7)& \hfill & \end{array}$$

Therefore, for all $x\in \mathbb{Z}$,

$$3x^2-x+5\equiv 0(\mathit{mod}7)⟺{(x+1)}^2\equiv 4(mod7).$$

Thus the solutions are $1,4$ modulo $7$.

(c)

$$\begin{array}{llll}\hfill 2x^2+7x-10& \equiv 2(x^2-2x-5)& \hfill & \\ \hfill & \equiv 2\left[{(x-1)}^2-6\right](\mathit{mod}11).& \hfill & \end{array}$$

Therefore, for all $x\in \mathbb{Z}$,

$$2x^2+7x-10\equiv 0(\mathit{mod}11)⟺{(x-1)}^2\equiv 6(mod11).$$

Since $6$ is not a square modulo $11$, this equation has no solution.

(d)

$$\begin{array}{llll}\hfill x^2+x+1& \equiv {(x+7)}^2-9(\mathit{mod}13)& \hfill & \end{array}$$

Therefore, for all $x\in \mathbb{Z}$,

$$x^2+x+1\equiv 0(\mathit{mod}13)⟺{(x+7)}^2\equiv 9(mod13).$$

Thus the solutions are $3,9$ modulo $13$.